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ANALYTIC GEOMETRY 



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JlfeQra&'j/illBoQk G±7n& 

PUBLISHERS OF &OOKS FOJO 

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ANALYTIC GEOMETRY 

WITH INTRODUCTORY CHAPTER ON THE 

CALCULUS 



BY 

CLAUDE IRWIN PALMER 

ASSOCIATE PROFESSOR OF MATHEMATICS, 
ARMOUR INSTITUTE OF TECHNOLOGY 

AND 

WILLIAM CHARLES KRATHWOHL 

ASSOCIATE PROFESSOR OF MATHEMATICS, 
ARMOUR INSTITUTE OF TECHNOLOGY 



First Edition 



McGRAW-HILL BOOK COMPANY, Inc. 
NEW YORK: 370 SEVENTH AVENUE 

LONDON: 6 6 8 BOUVERIE ST., E. C. 4 
1921 



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Copyright, 1921, by the 
McGraw-Hill Book Company, Inc. 



THK MAPI! PKBSS TOSKP1 



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V 



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r-i 



PREFACE 



The object of this book is to present analytic geometry to 
the student in as natural and simple a manner as possible 
without losing mathematical rigor. The average student 
thinks visually instead of abstractly, and it is for the average 
student that this work has been written. It was prepared 
primarily to meet the requirements in mathematics for the 
second half of the first year at the Armour Institute of Tech- 
nology. To make it adaptable to courses in other institutions 
of learning certain topics not usually taught in an engineering 
school have been added. 

While it is useless to claim any great originality in treat- 
ment or in the selection of subject matter, the methods and 
illustrations have been thoroughly tested in the class room. 
It is believed that the topics are so presented as to bring the 
ideas within the grasp of students found in classes where 
mathematics is a required subject. No attempt has been 
made to be novel only; but the best ideas and treatment have 
been used, no matter how often they have appeared in other 
works on the subject. 

The following points are to be especially noted: 

(1) The great central idea is the passing from the geometric 
to the analytic and vice versa. This idea is held consistently 
throughout the book. 

(2) In the beginning a broad foundation is laid in the 
algebraic treatment of geometric ideas. Here the student 
should acquire the analytic method if he is to make a success 
of the course. 

(3) Transformation of coordinates is given early and used 
frequently throughout the book, not confined to a single 
chapter as is so frequently the case. The same may be said 
of polar coordinates. 

v 



M305O99 



vi PREFACE 

(4) Fundamental concepts are dealt with in an informal 
as well as in a formal manner. The informal often fixes 
and clarifies the ideas where the formal does not. 

(5) Numerous illustrative examples are worked out in 
order that the student may get a clear idea of the methods 
to be used in the solution of problems. 

(6) The conic sections are treated from the starting point 
of the focus and directrix definition. 

(7) Because of its great importance in engineering practice 
the empirical equation is dealt with more completely than is 
usual. This treatment has been made as elementary as 
possible, but sufficiently comprehensive to enable one to 
solve the average problem in empirical equations. 

(8) The fundamental concepts of the calculus are presented 
in a very concrete manner, and a much greater use then is 
usual is made of the differential. The ideas are thus more 
readily visualized than is possible otherwise. The applications 
are mainly to tangents, normals, areas, and the discussion of 
equations. 

(9) The concluding chapter gives an adequate and careful 
treatment of solid geometry so necessary in the study of the 
calculus. 

(10) The exercises are numerous, carefully graded, and 
include many practical applications. 

(11) In the introductory chapter are found various short 
tables and formulas, and at the end are given four place 
tables of logarithms and trigonometric functions. 

The authors take this opportunity to express their indebted- 
ness to their colleagues, Professors D. F. Campbell, H. R. 
Phalen, and W. L. Miser, for their assistance in the preparation 
of the text. 

The Authors. 
Chicago, III., 
May, 1921. 



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CONTENTS 

CHAPTER I 
Introduction 

Abt. Pao« 

1. Introductory remarks 1 

2. Algebra and geometry united 1 

3. Fundamental questions 1 

4. Algebra • 2 

5. Trigonometry 3 

6. Useful tables 5 

CHAPTER II 
Geometric Facts Expressed Analytically, and Conversely 

7. General statement. 8 

8. Points as numbers, and conversely 8 

9. The line segment 9 

10. Addition and subtraction of line segments 10 

11. Line segment between two points 11 

12. Geometric addition and subtraction of line segments 11 

13. Determination of a point in a plane 12 

14. Coordinate axes 12 

15. Plotting a point 13 

16. Oblique cartesian coordinates 15 

17. Notation 15 

18. Value of a line segment parallel to an axis 16 

19 Distance between two points in rectangular coordinates ... 17 

20. Internal and external division of a line segment 19 

21. To find the coordinates of a point that divides a line segment in 

a given ratio 20 

22. Formulas for finding coordinates of point that divides a line 
segment in a given ratio 22 

23. The angle between two lines 24 

24. Inclination and slope of a line 25 

25. Analytic expression for slope of a line 25 

26. Formula for finding the slope of a line through two points . . 25 

vii 



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viii CONTENTS 

Abt. Pagb 

27. The tangent of the angle that one line makes with another in 
terms of their slopes 26 

28. Parallel and perpendicular lines 27 

29. Location of points in a plane by polar codrdinates 29 

30. Relations between rectangular and polar codrdinates 32 

31. Changing from one system of axes to another 33 

32. Translation of coordinate axes 33 

33. Rotation of axes. Transformation to axes making an angle <p 
with the original 34 

34. Area of a triangle in rectangular coordinates 36 

35. Area of any polygon 38 

36. Analytic methods applied to the proofs of geometric theorems 39 

CHAPTER III 
Loci and Equations 

37. General statement 44 

38. Constants and variables 44 

39. The locus 45 

40. The locus of an equation 45 

41. Plotting an equation 46 

42. The imaginary number in analytic geometry 47 

43. Geometric facts from the equation 48 

44. Intercepts • 49 

45. Symmetry, geometric properties 49 

46. Symmetry, algebraic properties 50 

47. Extent 51 

84. Composite loci 53 

49. Intersection of two curves 54 

50. Equations of loci 55 

51. Derivation of the equation of a locus 56 

CHAPTER IV 
The Straight Line and the General Equation of the First Degree 

52. Conditions determining a straight line 59 

53. Point slope form of equation of the straight line 59 

54. Lines parallel to the axes 60 

55. Slope intercept form 61 

56. Two point form 62 

57. Intercept form 62 



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CONTENTS ix 

Art. Pagb 

58. Normal form 63 

59. linear equations 65 

60. Plotting linear equations . . . . • 65 

61. Comparison of standard forms 66 

62. Reduction of Ax + By + C - O to the normal form .... 66 

63. Distance from a point to a line 68 

64. The bisectors of an angle 70 

65. Systems of straight lines 71 

66. Applications of systems of straight lines to problems 72 

67. Loci through the intersection of two loci 75 

68. Plotting by factoring 77 

69. Straight line in polar coordinates 78 

70. Applications of the straight line . 79 

CHAPTER V 
The Circle and Certain Forms of the Second Degree Equation 

71. Introduction 86 

72. Equation of circle in terms of center and radius 86 

73. General equation of the circle 87 

74. Special form of the general equation of the second degree . . 87 

75. Equation of a circle satisfying three conditions 88 

76. Systems of circles 92 

77. Locus problems involving circles 94 

78. Equation of a circle in polar coordinates 96 

CHAPTER VI 
The Parabola and Certain Forms of the Second Degree Equation 

79. General statement. 98 

80. Conic sections * . . 98 

81. Conies 99 

82. The equation of the parabola 100 

83. Shape of the parabola 101 

84. Definitions 102 

85. Parabola with axis on the t/-axis 102 

86. Equation of parabola when axes are translated 103 

87. Equations of forms y 2 + Dx + Ey + F = and x* + Dx + 

Ey +F = O 106 

88. The quadratic function ax 2 + by + c 107 

89. Equation simplified by translation of coordinate axes .... 107 



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X CONTENTS 

Abt. Pages 

90. Equation of a parabola when the coordinate axes are rotated 109 

91. Equation of parabola in polar coordinates Ill 

92. Construction of a parabola 112 

93. Parabolic arch 113 

94. The path of a projectile 114 

CHAPTER VII 
The Ellipse and Certain Forms of the Second Degree Equation 

95. The equation of the ellipse . 117 

96. Shape of the ellipse 119 

97. Definitions 120 

98. Second focus and second directrix 120 

99. Ellipse with major axis on the y-axis 121 

100. Equation of ellipse when axes are translated 123 

101. Equation of the form Ax 2 + Cy* + Dx + Ey + F - . . . 125 

102. Equation of ellipse when axes are rotated 127 

103. Equation of ellipse in polar coordinates 129 

104. Construction of an ellipse 129 

105. Uses of the ellipse 131 

CHAPTER VIII 

The Hyperbola and Certain Forms of the Second Degree 

Equation 

106. The equation of the hyperbola 134 

107. Shape of the hyperbola 136 

108. Definitions 137 

109. Second focus and second directrix 137 

110. Hyperbola with transverse axis on the y-axis 137 

111. Asymptotes 139 

112. Conjugate hyperbolas 141 

113. Equilateral hyperbolas 142 

114. Equation of hyperbola when axes are translated 143 

115. Equation of the form Ax* + Cy 2 + Dx + Ey + F = . . . 144 

116. Equation of hyperbola when axes are rotated 146 

117. Equation of hyperbola in polar coordinates 147 

118. Construction of an hyperbola. 148 

119. Uses of the hyperbola 150 



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CONTENTS Xi 



CHAPTER IX 
Other Loci and Equations 

Abt. Pao» 

120. General statement 164 

121. Summary for second degree equations 164 

122. Suggestions for simplifying second degree equations 167 

123. Parabolic type 158 

124. Hyperbolic type 159 

125. The cissoid of Diocles 160 

126. Other algebraic equations 161 

127. Exponential equations . 163 

128. Applications 164 

129. Logarithmic equations 165 

130. The sine curve 167 

131. Periodic functions 168 

132. Period and amplitude of a function 169 

133. Projection of a point having uniform circular motion. Simple 
harmonic motion 170 

134. Other applications of periodic functions 172 

135. Exponential and periodic functions combined 172 

136. Discussion of the equation • 174 

137. Loci of polar equations 175 

138. Remarks on loci of polar equations 177 

139. Spirals 178 

140. Polar equation of a locus 178 

vl41. Parametric equations 180 

142. The cycloid. . 182 

143. The hypocycloid 183 

144. The epicycloid 185 

145. The involute of a circle 186 



CHAPTER X 

Empirical Loci and Equations 

146. General statement 188 

147. Empirical curves 188 

148. Experimental data 190 

149. General forms of equations . K 191 

150. Straight line, y = mx + b 191 

151. The method of least squares 193 



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xii CONTENTS 

Abt. Paob 

152. Parabolic type, y = ex*, n > 195 

153. Hyperbolic type, y = cx n , n < 197 

154. Exponential type, y - ab* or y = oe** 197 

155. Probability curve 199 

156. Logarithmic paper 200 

157. Empirical formulas of the type y *= a + bx + ex* + dx 8 + gx n 203 

CHAPTER XI 
Poles, Polars, and Diameters 

158. Harmonic ratio 206 

159. Poles and polars 206 

160. Properties of poles and polars 209 

161. Diameters of an ellipse 210 

162. Conjugate diameters of an ellipse 212 

163. Diameters and conjugate diameters of an hyperbola .... 213 

164. Diameters and conjugate diameters of a parabola . . . 213 

165. Diameters and conjugate diameters of the general conic . . . 214 

CHAPTER XII 
Elements op Calculus 

166. Introductory remarks 216 

167. Functions, variables, increments 216 

168. Illustrations and definitions 220 

169. Elementary theorems of limits 221 

170. Derivatives 222 

171. Tangents and normals 223 

172. Differentiation by rules 225 

173. The derivative when J (x) is x 226 

174. The derivative when f(x) is c 226 

175. The derivative of the sum of functions 226 

176. The derivative of the product of two functions 227 

177. The derivative of the product of a constant and a function 227 

178. The derivative of the quotient of two functions 228 

179. The derivative of the power of a function 228 

180. Summary of formulas for algebraic functions 230 

181. Examples of differentiation 231 

182. Differentiation of implicit functions 233 

183. Discussion of uses of derivative 235 

184. Properties of a curve and its function 235 



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CONTENTS xiii 

Art. Pagb 

185. Curves rising or falling, functions increasing or decreasing 236 

186. Maximum and minimum 237 

187. Concavity and point of inflection 239 

188. Relations between increments 242 

189. Differentials 243 

190. Illustrations 243 

191. The inverse of differentiation 246 

192. Determination of the constant of integration 247 

193. Methods of integrating 248 

194. Trigonometric functions 250 

195. Derivatives of sin u and cos u 250 

196. Derivatives of other trigonometric functions 252 

197. f sin udu and f cos udu 253 

198. Derivative of log«i* 254 

199. Derivative of log w ' 255 

200. Derivative oLa* and e u 255 

201. Derivative of -u* 256 

202. Illustrative examples 257 

203. f^,f e u du, and f a u du 258 

CHAPTER XIII 
Solid Analytic Geometry 

204. Introduction .... 261 

205. Rectangular coordinates in space 261 

206. Geometrical methods of finding the coordinates of a point in 
space 263 

207. Distance between two points 263 

208. Coordinates of a point dividing a line segment in the ratio 
fito r« 264 

209. Orthogonal projections of line segments 266 

210. Direction cosines of a line 267 

>211. Polar coordinates of a point 269 

212. Spherical coordinates 270 

213. Angle between two lines 271 

214. Locus in space 274 

215. Equations in one variable. Planes parallel to the axes .... 274 

216. Equations in two variables. Cylindrical surfaces 274 

217. Spheres 276 

218. Surfaces of revolution 276 

219. Equations of curves in space 278 



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xiv CONTENTS 

Abt. Paob 

220. Sections of a surface by planes parallel to the coordinate planes. 279 

221. Projections of curves on the codrdinate planes 280 

222. Surfaces in space 282 

223. General equation of second degree 284 

224. Ellipsoid 284 

226. The hyperboloid of one sheet 285 

226. The hyperboloid of two sheets 287 

227. Elliptic paraboloid 288 

228. Hyperbolic paraboloid 289 

229. Cone 290 

230. Equation of a plane 292 

231. General equation of a plane 292 

232. Normal form of the equation of a plane 293 

233. Reduction of the equation of a plane to the normal form. . . 293 

234. Intercept form of the equation of a plane 294 

235. The equation of a plane determined by three conditions . 295 

236. Angle between two planes 295 

237. Distance from a point to a plane 296 

238. Two plane equation of a straight line 298 

239. Projection form of the equation of a straight line 298 

240. Point direction form of the equation of a straight line, 
symmetrical form 299 

241. Two point form of the equation of a straight line 300 

Summary of Formulas 303 

Four Place Table op Logarithms 308 

Table of Trigonometric Functions 310 

Answers 315 

Index 341 



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ANALYTIC GEOMETRY 

CHAPTER I 
INTRODUCTION 

1. Introductory remarks. — Although it is not always possi- 
ble for a student to appreciate at the outset the content of a 
subject, it is well, however, to consider the object of the study, 
and to understand as far as possible its fundamental aims. 

2. Algebra and geometry united. — Analytic geometry, or 
algebraic geometry, is a subject that unites algebra and geom- 
etry in such a manner that each clarifies and helps the other. 
Lagrange says: "As long as algebra and geometry travelled 
separate paths their advance was slow and their applications 
limited. But when these two sciences joined company, they 
drew from each other fresh vitality and thenceforward marched 
on at a rapid pace towards perfection. It is to Descartes 1 
that we owe the application of algebra to geometry — an appli- 
cation which has furnished the key to the greatest discoveries 
in all branches of mathematics." 

3. Fundamental questions. — The fundamental questions of 
analytic geometry are three. 

First, given a figure defined geometrically, to determine its 
equation, or algebraic representation. 

1 Rene* Descartes (1596-1650) was one of the most distinguished philos- 
ophers. It was in pure mathematics, however, that he achieved the 
greatest and most lasting results, especially by his invention of analytic 
geometry. In developing this branch he had in mind the elucidation of 
algebra by means of geometric intuition and concepts. He introduced 
the present plan of representing known and unknown quantities, gave 
standing to the present system of exponents, and set forth the well 
known Descartes' Rule of Signs. His invention of analytic geometry 
may be said to constitute the point of departure of modern mathematics. 

1 



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2 ANALYTIC GEOMETRY [§4 

Second, given numbers or equations, to determine the geo- 
metric figure corresponding to them. 

Third, to study the relations that exist between the geo- 
metric properties of a figure and the algebraic, or analytic, 
properties of the equation. 

To pursue the subject of analytic geometry successfully the 
student should be familiar with plane and solid geometry, and 
should know algebra through quadratic equations and plane 
trigonometry. 

While parts of analytic geometry can be applied at once to 
the solution of various interesting and practical problems, 
much of it is studied beeause it is used in more advanced 
subjects in mathematics. 

Some of the more frequently used facts of algebra and trig- 
onometry are given here for convenience of reference. 

4. Algebra. — Quadratic equations. — The roots of the quad- 
ratic equation ax 2 + bx •+ c = are 

—b + Vb 2 - 4ac , m —b — y/b 2 — 4ac 

n = — • 7p , and r 2 = = 

2a 2a 

n + f2 = 9 and rif2 = — 

a a 

These roots are 

real and equal if 6 2 — 4ac = 0, 
real and unequal if b 2 — 4ac>0, 
imaginary if 6 2 — 4ac<0. 

The expression b 2 — 4ac is called the discriminant of the 
quadratic equation. 
Logarithms. 

(iy log M N = log M + log N. 

(2) log (M -5- N) = log ilf - log JV. 

(3) log N n = n log AT. 



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§5] INTRODUCTION 3 

(7) a 10 *" = N. 

(8) log ^ = -log AT. (10) log* clog. 6 = 1. 

" j (11) log, N = 2.302585 log i0 N. 

(9) log* N = j^-£ log. N. (12 ) i ogl0 N = 0.43429 log. N. 

The base e = 2.718281828459- • -. t = 3.141592653589- • •. 
6. Trigonometry. — Formulas. 

(1) 2t radians = 360°, «■ radians = 180°. 

lSfl° 

(2) 1 radian = — = 57.29578° - - 57° 17' 44.$". 

IT 

(3) 1° = j|q = 0.0174533 - radians. 

(4) sin 2 + cos 2 = 1. 

(5) 1 + tan 2 = sec 2 0. 

(6) 1 + cot 2 = esc 2 0. 
1 , „ 1 



(7) sin = ^, and esc = . _ 

v ' esc 0' sin 

(8) cos = — £, and sec = 



sec 0' cos 

(9) tan = —r-z y and cot = z — 5 - 
v ' cot0 tan0 

(10) tan = - = s - 

v ' COS0 C8C0 

/ 11N , - COS0 CSC0 

(11) COt = - — ^ = ■=• 

' sin sec 

(12) sin (a + B) = sin a cos B + cos a sin B. 

(13) cos (a + B) = cos a cos B — sin a sin 0. 

(14) sin (a — B) = sin a cos B — cos a sin 0. 

(15) eos (a — B) = cos a cos B + sin a sin 0. 

/i£!\ x / i o\ tan a + tan 

(16) tan (a + B) = = t r — ^* 

v ' 1 — tan a tan B 

fi*\ x / x>\ tana — tan £ 

(17) tan (cr - 0) = r - 1 — — -^=. — * 

1 + tan a tan 

(18) sin 20 = 2 sin cos 0. 

(19) cos 20 = cos 2 - sin = 1 - 2 sin 2 = 2 cos 2 0-1. 
/oa\ x o/i 2 tan 

(20)ten2 ' = l-^taF? 

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ANALYTIC GEOMETRY 



[(5 



(21) sin \0 



t 



— C08 



(23) tan 



*--±>/nF 



(22) cos 
— cos 1 — cos 



»-±£ 



+ cos 6 



2 

sin 



(24) 
(25) 
(26) 
(27) 
(28) 
(29) 
(30) 
(31) 

(32) 

(33) 
(34) 



(35) 



(36) 



(37) 



(38) 



cos B sin B 1 + cos B 

sin a + sin = 2 sin J(a + 0) cos J(a — 0). 
sin a — sin = 2 cos J(a + 0) sin J(a — 0). 
cos a + cos = 2 cos J(a + 0) cos J(a — 0). 
cos a — cos = —2 sin J(a + 0) sin \{a — 0). 
sin a cos = J sin (a + 0) + i sin (a — 0). 
cos a sin = \ sin (a + 0) — J sin (a — 0). 
cos a cos = J cos (a + 0) + J cos (a .— 0). 
sin a sin = —J cos (a + 0) + J cos (a — 0). 

(Sine Law.) 



(J* — 0) = cos 0. 
(§* — *)" sin •• 
(£* - 0) m cot 0. 
( Jt - 0) = tan 0. 
(§* + •) — cos 0. 

(§* + •) = —si 11 •• 
(Jr + 0) = -cot0. 

(J* + •) = -tan0. 
( * — 0) = sin0. 
( ir — 0) = —cos 0. 
( it — 0) « — tan0. 
( ir — 0) = — cot0. 
( ir + 0) = — sin0. 
( *■ + 0) = — COS0. 
( ir,+ 0) = tan0. 
( ir + 0) = cot 0. 
(■§tt — 0) = — cos0. 
(fir - 0) = -sin0. 
(|tt — 0) = cot 0. 
($* - 0) » tan 0. 



(Cosine Law.) 



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§6) 



INTRODUCTION 



(39) sin (frc + 0) « -cos 0. 
cos (-fir + 0) = sin 0. 
tan (fir + 0) = — cot0. 
cot (fir + 0) = — tan0. 

(40) sin (2ir - 0) = -sin 0. 
cos (2* — 0) = cos 0. 
tan(2ir — 0) = — tan0. 
cot (2* - 0) = -cot0. 

(41) sin ( — 0) = — sin 0. 
cos ( — 0) = cos 0. 
tan ( — 0) = — tan 0. 
cot ( — 0) = — cot0. 

6. Useful tables. 







Values 


of e* 


FROM 


x -0 


TO X 


-4.9 






X 


0.0 


0.1 


0.2 


0.3 


0.4 


0.5 


0.6 


0.7 


0.8 


0.9 





1.00 


1.11 


1.22 


1.35 


1.49 


1.65 


1.82 


2.01 


2.23 


2.46 


1 


2.12 


3.00 


3.32 


3.67 


4.06 


4.48 


4.95 


5.47 


6.05 


6.69 


2 


7.39 


8.17 


9.03 


9.97 


11.0 


12.2 


13.5 


14.9 


16.4 


18.2 


3 


20.1 


22.2 


24.5 


27.1 


30.0 


33.1 


36.6 


40.4 


44.7 


49.4 


4 


54.6 


60.3 


66.7 


73.7 


81.5 


90.0 


99.5 


109.9 


121.5 


134.3 





Values of t 


r* FROM X 


= TO X = 


4.9 






X 


0.0 


0.1 


0.2 


0.3 


0.4 


0.5 


0.6 


0.7 


0.8 


0.9 





1.00 


0.90 


0.82 


0.74 


0.67 


0.61 


0.55 


0.50 


0.45 


0.41 


1 


0.37 


0.33 


0.30 


0.27 


0.25 


0.22 


0.20 


0.18 


0.17 


0.15 


2 


0.14 


0.12 


0.11 


0.10 


0.09 


0.08 


0.07 


0.07 


0.06 


0.06 


3 


0.05 


0.05 


0.04 


0.04 


0.03 


0.03 


0.03 


0.02 


0.02 


0.02 


4 


0.02 


0.02 


0.01 


0.01 


0.01 


0.01 


0.01 


0.01 


0.01 


0.01 



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ANALYTIC GEOMETRY 



[§6 





?H 




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1-4 


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1 

04 

1 


OQ 


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04 

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csc 



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§6] ^■to^INTRODUCTION 7 

Table of Frequently Used Trigonometric Functions 



6° 


in 
radians 


sin0 


cos 


tan 


cot 


sec 


CSC $ 


0° 








1 





00 


1 


00 


30° 


5 


1 
2 


V3 
2 


. V3 
3 


V5 


2\/5 
3 


2 


45° 


4 


V2 
2 


2 


1 


1 


V5 


V5 


60° 


5 


V3 
2 


1 
2 


V3 


3 


2 


2V3 
3 


90° 


2 


1 





00 





00 


I 


120° 


2t 
3 


V3 
2 


1 
2 


-V5 


V3 
3 


-2 


2VS 

3 


135° 


3t 
4 


V2 
2 


V2 
2 


-l 


-1 


-V5 


V5 


150° 


5t 


1 


V3 


V3 


-V3 


2VS 


2 




6 


2 


2 


3 


3 




180° 


T 





-1 





00 


-1 


00 


210° 


7x 
6 


1 
2 


V3 
2 


V3 
3 


vs 


2\/3 
3 


-2 


225° 


5t 
4 


V2 
2 


V2 
2 


1 


1 


-V5 


-V5 


240° 


4x 
3 


V3 
2 


1 
2 


V3 


3 


-2 


2\/3 
3 


270° 


3t 
2 


-1 





00 





00 


-1 


300° 


5t 
3 


V3 
2 


1 
2 


-V3 


V3 
3 


2 


2\/3 
3 


315° 


7x 
^4~ 


V2 
2 


V2 
2 


-1 


-1 


VS 


-V2 


330° 


llx 
6 


1 
2 


V3 
2 


V3 
3 


-V3 


2a/3 
3 


-2 


360° 


2x 





1 





00 


1 


00 



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CHAPTER II 

GEOMETRIC FACTS EXPRESSED ANALYTICALLY, AND 
CONVERSELY 

7. General statement. — Geometry deals with points, lines, 
and figures composed of points and lines. Algebra deals with 
numbers and algebraic statements composed of numbers, such 
as the equation. 

In order to study geometric relations by means of algebra, 
and conversely, it is necessary to be able to represent points, 
lines, and geometric figures by means of numbers and equa- 
tions, and conversely. That is, it is necessary to be able to 
translate from the language of geometry to that of algebra, and 
conversely. 

8. Points as numbers, and conversely. — If a point moves 
from A to £ in a straight line, the point is said to generate 
the line segment AB, that is, the line segment AB. is the locus 
of the point. If the point moves from B to A it generates 

the line segment BA. It is con- 

— ■ — *? venient to consider AB and BA as 

Pjq x separate line segments having oppo- 

site directions. The arrow is often 
used to denote the positive direction. 

Such line segments as AB and BA are called directed line 
segments. The point from which the moving point starts is 
called the initial point, and the point where it stops is called 
the terminal point. 

It is to be noted that a line segment is read by naming the 
initial point first. 

Let X'X be a straight line of indefinite length, and 
choose: first, a unit of length; second, a direction of motion, 
which we shall call positive if toward the right and negative 

8 



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§9] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 9 

if toward the left; third, a point called the origin from which 
to start. 

Then any poiqt P can be determined by a real number — inn 
tegral, fractional, or irrational — which shows the number of 
units the point has moved from the origin. 

The number is positive or negative according as the motion 
is in the positive or negative direction. The origin is desig- 
nated by 0. 

Conversely, any real number corresponds to a point which 
is distant that number of units in the proper direction from 
the origin. 

Thus, in Fig. 2, +6 designates the point Pi and —2 the point Pi; 
while R corresponds to — 3^, and Q to V3. 

The line X f X is a directed straight line if it is thought of as 
generated by a point moving in the direction from X 9 to X or 
from X to X'. 

.Unit. 



_/ , , J* Pt O Q P P, 

*— j 1 J—M 1 1 1 1 H 1 ^— H H*X 

-6-6-4-3-2-1 1^2 3 4 5 « 

Fig. 2 

9. The line segment. — The magnitude of a line segment is 
determined by the number of units in its length, that is, by 
the number of units a point moves in generating it. 

The value of a line segment is determined by its length and 
direction, and is defined to be the number which would represent 
the terminal point of the segment if the initial point were taken 
as origin. 

It follows from this definition that the value of a line seg- 
ment read in one direction is the negative of the value if read 
in the opposite direction. 

Thus, AB - -BA, or AB + BA - 0. 

By the numerical value of a line segment is meant the num- 
ber of units of length in it without reference to its direction. 

Two line segments are equal if they have the same direction 
and the same length, that is, the same value. 



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10 ANALYTIC GEOMETRY [§10 

In Pig. 3, AB - +2, CD - +2, ZW - +6, EC « -4, FA - -8„ 
AB — CD, and AF — CAT. AC and FD are equal in numerical value. 

A BC i> # F N , Unit 

t ♦— 1 * h-H* — » H » 1 1 H- *- »— I 

Fig. 3. 

EXERCISES 

1. Draw a line segment 5 in. long and take the origin at the center. 
Choose as a unit of measure a line i in. long. What numbers designate 
the ends of the line? Locate the points corresponding to the numbers 9, 
7i -4, -3i V2, - aA-t. 

2. Draw a line segment 20 units in length, with the origin, 0, at the 
center. Locate the following points: A corresponding to 3, B corre- 
sponding to 8, C corresponding to — 4, D corresponding to — 10, E cor- 
responding to 10. Give the values of the following line segments: AB, 
DA, CE, BC, EA, AC. 

3. In exercise 2, how are the numbers designating the points affected 
if the origin is moved two units to the right? How are the values of the 
line segments affected? 

L A C B M iy w 

% 1 1 1 1 1 H>- 

Fig. 4. 

10. Addition and subtraction of line segments. — In Fig. 4, 
if A, B, C, • • • M, N are any arrangement of points on a 
^ p P straight line, then 

•"£ ^ f AB + BC+ • - - +MN + NA-0. 

Pl Pt For the moving point generates in succes- 

* ' *" sion the line segments AB, BC, • • • MN, 

IG * NA, starting at A and returning to A. It 

therefore generates as much in the negative direction as in the 

positive. Hence the sum is zero. 

A case of frequent occurrence is that of three points 0, Pi, 
and P2 on a straight line, Fig. 5. If is taken as origin, then 

[1] (1) OP 2 = OPi + PxP2, 

(2) PJ> 2 - OP 2 - OPi. 



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§11] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 11 

Proof. OPi + P X P 2 + P2O = 0. Why? 
Adding OP 2 = OP 2 , gives OP x + P1P2 = OP*. 
Adding OP 2 + PiO = OP* + PiO, gives PiP* =[OP 2 + P x O. 
.'. PiP 2 = OP 2 - OPu 

11. Line segment between two points.— To find the value 
of the line segment between two points on a straight line, when 
the numbers determining these two points with reference to 
an origin on the same line, are known. 

In Fig. 5, is the origin and x x and x 2 are the numbers 
determining the points Pi and P 2 respectively. It is required 
to find the value of the line segment P1P2, that is, the magni- 
tude and direction of P1P2. 

P1P2 = OP 2 - OPi. By [1]. 

But OPi = xi, OP 2 = x 2 . 

[2] .*. P1P2 = x 2 - xi. 

This states that the value of the line segment between two points 
on a straight line is equal to the number determining the terminal 
point of the line segment minus the number determining its initial 
point, when a point on the straight line is taken as the origin. 

p 6 p 6 O Pi Pi Pz A 

1 i 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ' 1 1 ' p 

-11-10-9 -8-7-6-5-4-3-2-1 1 2 » 4 5 7 8 9 10 U 12 13 
FlQ. 6. 

Thus, in Fig. 6, PiP% = OP» - OP x - 5 - 3 = 2. 
PJPt = OPi - OP, = 5 - 8 - -3. 
PsPe - OP, - OP, = -4 - (-8) - +4. 
PJ>t - OPt - OP, - 5 - (-4) - +9. 
PzP, - OP, - OP, - -8 - 8 - -16. 

12. Geometric addition and subtraction of line segments. 
From the preceding article it readily follows that two line seg- 
ments having the same or opposite directions can be added by 
placing the initial point of the second upon the terminal point 
of the first. The sum of the line segments is the line segment 



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12 ANALYTIC GEOMETRY [§13 

having, as initial point, the initial point of the first and } as terminal 
point, the terminal point of the second. 

A line segment is subtracted from another by reversing its 
direction and adding. 

Thus, in Fig. 6, OP x + PiP, - OP* - 8. 
Pft + PJ>> - PJ>> « -11. 
PJ>, - PtP* - PJ> % + PJ> t - PtPi - 9. 
PJ>< - P.P4 - PJ>* + P4Pt - PiPt - -9. 

EXERCISES 

1. On a line with origin at 0, locate the following points: A determined 
by 2, B by 3, C by 8, Dby -5, E by -8. By the method of article 10, 
find the value of the line segments AB, BC, BD, AE, DE, EB, CE, CD. 

2. On a line with origin at 0, locate the points Pi, P s , P s , Pa, determined 
by the numbers x x , x ty x 9 , x* respectively. (1) Give the values of the 
line segments PiP», PiP*, PiPj, P4PL (2) Give the line segments that 
have the following values: x* — x h X\ — Xt, x% — x t . Do the relative 
positions of the points make any difference in the answers? 

13. Determination of a point in a plane. — It was shown in 
article 8 that the position of a point on a straight line can be 
determined by one number, which shows the direction and the 
distance that the point is from a fixed point on the straight line. 

Various methods may be given for locating a point in a plane. 
For the purposes of analytic geometry, two of these will be 
chosen. They correspond to the two methods ordinarily used 
in locating a point on the surface of the earth. 

First, a house in a city is located by giving its street and 
number. That is, by stating its distance and direction from 
each of two intersecting streets. 

Second, a city may be located by giving its distance and 
direction from another city. 

In analytic geometry, the two corresponding methods of 
locating a point in a plane are (1) the method by cartesian 
codrdinates, and (2) the method by polar coordinates. 

CARTESIAN COORDINATES 

14. Coordinate axes. — (1) The lines of reference X'X'and 
Y'Y, Fig. 7, intersecting in the point 0, are chosen. These 



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§15] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 13 

lines are considered perpendicular to each other in this article, 
and will always be so taken unless otherwise stated. 

The line X'X is called the axis of abscissas or the x-axis. 
The line Y f Y is called the axis of ordinates or the y-axis. To- 
gether they are called the coordinate axes. 

When the coordinate axes are perpendicular to each other 
they form a rectangular system. 

The coordinate axes divide the plane into four quadrants, 
numbered I, II, III, and IV as in trigonometry. 

(2) A line segment of convenient length is chosen for a unit 
of measure. This may be of any length whatever. 

(3) The direction is chosen as positive when towards the 
right parallel to the x-axis, or 
upwards parallel to the y-axis. p f 
Hence the negative direction 
is towards the left, or down- 
wards. 

15. Plotting a point.— A x'-+ 
point Pi in the plane is de- 



*L 



M % O 



*i 



Nt 



*** 



*P* 



Fio. 7. 



termined by the line segments 
NiPi and Af 1P1, Fig. 7, drawn 
parallel to X'X and Y' Y re- 
spectively, for the values of 
these line segments tell how far and in what direction Pi is 
from the lines of reference. 

Here the line segment NiPi — +5, and M\P\ = +4. 

The point P% is determined by the line segments NtPi — — 3, and 
M,P S - +6. 

The point P 9 is determined by the line segments NtP 9 — — 6, and 
MsP, - -4. 

It is evident that any point in the plane is determined by 
one pair of numbers, and only one; and, conversely, every pair 
of real numbers determines one point in the plane, and only 
one. 

The two numbers that determine a point in a plane are 



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14 



ANALYTIC GEOMETRY 



1515 



palled the coordinates of the point. The number which is the 
value of the line segment parallel to the z-axis is called the 
abscissa of the point, and is usually represented by x. The 
number which is the value of the line segment parallel to the 
2/-axis is called the ordinate of the point, and is usually repre- 
sented by y. 

The coordinates are written, for brevity, within parentheses 
and separated by a comma, the abscissa always being first, as 
(x, y). The letter designating the point is often written just 
before the parentheses. 

Thus, the points in Fig. 7 are written: Pi(5, 4), P*( -3, 6), P»(-6, -4), 
and p4(8, —3). The points Mi, Mi, Ni, N t , and are respectively the 
points (5, 0), (-3, 0), (0, 4), (0, -4), and (0, 0). 

It is evident that, in the first quadrant, both coordinates 
are positive; in the second quadrant, the abscissa is negative 
and the ordinate positive; in the third quadrant, both 
coordinates are negative; and, in the fourth quadrant, the 
abscissa is positive and the ordinate negative. 

When a point is located in a plane by means of its coordi- 
nates it is said to be plotted. 

The locating of points is 
v greatly facilitated by using 

paper that is ruled into small 
squares. Such paper is called 
coordinate paper. 





1 






4 


t 








1 




JJH 


4> 










1 


D 1< 


5,3;) 




















































































o 


































































p ? ( - t 


M) 








p,< 


B? 


h) 






\ ' 


1 








1 





Example. — Plot the points Pi(5,3), 
P»(5, -3), P,(-2, -4), andP 4 (-4,4). 
The point Pi(5, 3) is plotted by 
counting off from along X'X a num- 
*ber of divisions equal to the abscissa 
5, and then from the point so de- 
termined, a number of divisions on a 
line parallel to the ^-axis, equal to the ordinate 3. 

The points (5,-3), (-2,-4), and (-4, 4) are located in a similar 
manner. 



Y 

Fig. 8. 



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++~x 



§16] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 15 

16. Oblique cartesian coordinates. — In determining a point 
in a plane, it is not necessary that the coordinate axes shall 
be perpendicular to each other, but they may form an angle 
a). Such a set of axes is called an oblique cartesian system. 

In Fig. 9, the abscissa of Pi is NiPi - 3, and its ordinate is M iPi - 5. 
The coordinates of P t are N*Pt = — 4, and MiPj = 3. 

17. Notation. — To secure 
clearness of statement, sub- 
scripts will be used with the 
letters designating points, and 
they will agree with the sub- 
scripts used with the coordi- 
nates of the points. 

Thus, the point Pi has coordi- 
nates (x h j/i), the point P s has 
coordinates (xt, y»), and so on. 

Points designated in this 
manner will, in general, be fixed points, while a point that 
may vary in position will be designated by a letter, as P, 
without a subscript and have coordinates (rr, y). 

EXERCISES 

1. Draw a pair of axes and plot the following points:, (2, 3), (7, 9), 
(-2, 4), (-7, -2), (4, -3), (-2, -8), (0, 0), (0, 5), (-6, 0). 

2. Draw the triangle whose vertices are (0, 2), (—2, —3), and (3, —2). 

3. Draw the quadrilateral whose vertices are (3, 0), (0, 2), (— 6> 2), 
and(0, -2). 

4* If the ordinate of a point is 0, where is the point? Where if its 
abscissa is 0? Using x for the abscissa and y for the ordinate, express 
each as an equation. 

6. What is the locus of all points that have abscissas equal to 5? 
Of all points having ordinates equal to 10? Use x for the abscissas and 
y for the ordinates and write these statements as equations. 

6. The abscissas of two points are each a. How is the line joining 
them situated with reference to the y-axis? The ordinates of two 
points are each —6. Sow is the line joining them situated with reference 
to the x-axis? Write each of these lines as an equation. 

7. Two points are placed so that the abscissa of each is equal to 



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16 ANALYTIC GEOMETRY [§18 

its ordinate: How is the line joining the points situated with reference 
to the coordinate axes? In what two quadrants can the points lie? 
Write the equation. 

8. Two points are placed so that the abscissa of each is equal to 
the negative of the ordinate. How is the line connecting them situated 
with reference to the coordinate axes? In what two quadrants can the 
points lie? Write the equation. 

9. Draw a rectangle whose vertices are (—4, 2), (—4, —5), (7, —5), 
and (7, 2). Find the length of its sides by differences of abscissas or 
ordinates. 

10. The vertex of a square is at the origin, and a diagonal lies on the 
positive part of the x-axis. # Find the coordinates of the other vertices 
if a side is 10. 

11. What is the locus of a point which moves so that the ratio of 
its ordinate to its abscissa is always 1? So that this ratio is always — 1? 
Always 2? Write the equations. 

12. An equilateral triangle of side a has a vertex at the origin and 
one side on the x-axis at the right of the origin. Find the coordinates 
of its vertices. 

13. A regular hexagon of side 8 is placed so that its center is at the 
origin and one diagonal is along the x-axis. Find the coordinates of its 
vertices. 

18. Value of line segment parallel to an axis. — If the 
segment of a line is parallel to one of the coordinate axes, it 
has a definite direction as well as a length, that is, it has a 
value. If P\{x\> yi) and P%(x 2 , y 2 ) are any two points on a 
line parallel to the x-axis, then 

[2i] P1P2 = x 2 - Xi. 

This follows directly from article 11, for if P1P2 intersects 
the y-axis in JYi, P1P2 = NiP 2 — N1P1 = x 2 — x\. 

Likewise, if PiOd, y{) and P 2 (x 2) 2/2) are any two points on 
a line parallel to the y-axis, then 

[2 2 ] P1P2 = y 2 - yi. 

The student should locate points in various positions and 
satisfy himself that [2J and [2 2 ] are true. Figure 10 shows 
several positions of Pi and P2 

These facts may be stated as follows: 



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p* 


W k Pi 








Ni 


Pi 


r* 


/ 








— *x 



§19] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 17 

(1) The value of a line segment parallel to the x-axis equals 
the abscissa of its terminal point minus the abscissa of its initial 
point. 

(2) The value of a line segment parallel to the y-axis equals 
the ordinate of its terminal point minus the ordinate of its initial 
point. 

19. Distance between two points in rectangular coordi- 
nates. — (1) The distance between two points is the numerical 
value of the line segment connecting these points, that is, it is 
the length of the line segment connecting the two points. 
It follows that the distance between 
two points, having abscissas X\ and 
x% on a line parallel to the 3-axis, is 
either x* — x\ or x\ — X2 f the differ- 
ence being taken positive when its^l 
numerical value can be determined. 

Likewise, when the two points p, p l 

are on a line parallel to the y-axis 

the distance between them is y 2 — FlG> 10 

yioryi - y 2 . 

(2) Ordinarily a line segment that is not parallel to one of thfe 
coordinate axes does not have a direction assigned to it. We 
do not then speak of its value. The length of such a line 
segment is the distance between its end points. 

The distance between two points Pi(xi, y{) and P2O&2, yt) 
is given by the formula 
[3] d - V(x!-x 2 ) 2 + (yi-y2) 2 . 

Proof. — Let Pi(x h yi) and P 2 (x 2j 2/2) be the two points. 

Through Pi and P2 draw lines parallel, respectively, to 
the x-axis and y-axis to intersect in Q. 

Then P1QP2 is a right triangle, and 

d = PJ> 2 = Vp&* + QP a *. 
But PiQ = x 2 - x h and QP 2 = 2/2 - 2/1. By [2i] and [2 2 ]. 
Hence d = V(x 2 - *i) 2 + (2/2 - 2/1) 2 . 



ATi 



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18 



ANALYTIC GEOMETRY 



[510 



Since (x 2 - Xi) 2 = (xi - x 2 ) 2 and (y 2 - j/i) 2 = (y x - j/ 2 )*, 

d = V(xi - x 2 ) 2 + (2/1 - y*)*. 

It should be noted that the line through Pi could as well 

have been drawn parallel to the 
Jt(*!.v») y-axis, and the line through P* 
parallel to the rr-axis. 

It is to be noted that the above 
proof is general and is made with- 
out reference to a figure. The 
student, however, should draw 
several figures locating the points 
in different positions and satisfy 
[3]. Figure 11 shows one position of 



^ 



Pi(*i.*i) 



+X 



»i-*i Q(*«,lfi) 



Fig. 11. 



himself of the truth of 
the points. 



EXERCISES 



1. Find the distance between each of the following pairs of points: 

(1) (3, 4), (-6, -8). (3) (-1, 0),(12, -2). 

(2) (-10, 4), (3, -9). (4) (6, 7), (-5, -5). 



*i(«**i> 



2. In Fig. 12, express each of the fol- 
lowing line segments as the difference 
between two abscissas: M1M4, M*M* y 

3. Express each of the following line 
segments as the difference between two 
ordinates: NiN h N*N h NJfi, NtN*. 

4. Derive the formula for the distance 
between Pi(x h y x ) and Pi(xt t y»), (1) 
when both Pi and Pi are in the first 
quadrant, (2) when Pi is in the third 
and Pj in the fourth quadrant, (3) when 
Pi is in the fourth and P* in the second quadrant. 

6. Find the lengths of the sides of the following triangles: 
(1) (2, 3), (-5, 8), (-2, -4). (2) (3, -6), (0, 5), (-4, -2). 

6. Show that the points (9, 12), (-3, -4), and (5, 4 - 4\/§) lie on a 
circle whose center is at the point (3, 4). 

7. Find a point whose abscissa is 3 and whose distance from (—3, 6) 
is 10. 




^ (**.»♦) 



Fig. 12. 



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520] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 19 

Suggestion. — Let y be the ordinate of the point. 

Then V(3 + 3)* + (y - 6)» - 10. Solve for y. 

8. Find the center of the circle passing through the three points 
(6, 15), (13, 8), and (-4, -9). 

Suggestion. — By the definition of a circle, if a circle passes through 
these three points, there must be a point (x, y) from which they are 
equally distant. Write the distance of each point from the point (x, y) 
and form two equations. Solve these equations for x and,y. 

9. Three vertices of a parallelogram are (—2, 4), (5, 2), and (6, 1). 
Find a fourth vertex. How many are there? 

Suggestion. — Use the fact that the opposite sides of a parallelogram 
are equal. 

i 10. Two vertices of an equilateral triangle are (2, 10) and (8, 2). 
Find the third vertex. 

11. Find the equation which states that the point (x, y) is 5 units from 
the point (3, 4). What is the locus of the point (x, y)? Draw the 
locus. 

12. Find the equation that expresses the fact that the point (x, y) is 
equally distant from the points (2, 3) and (7, —4). What is the locus? 

13. Show that the values of line segments parallel to either axes in 
rectangular coordinates hold true when the axes are oblique. 

14. If the axes are inclined to each other at an angle of <•>, and if 
lines PiQ and QP t of Fig. 11 are drawn parallel to the axes, then the 
angle P\QP% equals <•> or 180° — w. By the cosine theorem of t igonom- 
etry show that then the distance between the two points Pi(x\, yQ and 
P«(*i, Vt) is d = V(*i - xt) 1 + (yi - y t )* + 2{x x - x t )(yi - y%) cos «. 

16. The angle between two oblique axes is 60°. Find the distance 
between the points (—2, 3) and (6, —4). 

DIVISION OF A LINE SEGMENT 

20. Internal and external division of a line segment. — 
If Pi and P% are any two points on a straight line, then any 
third point, P , on the line is 
said to divide the line seg- -& ? l ^° ?* p ° 



H h- 



ment PiP 2 into two parts. Fjq 13 

The point P is said to 
divide the line segment PiP 2 internally if P lies between Pi 
and P 2 ; and externally if P lies beyond P 2 as at P , or be- 
yond Pi as at Pj. 



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20 ANALYTIC GEOMETRY [§21 

When P lies between Pi and P* the two parts are P1P0 and 
PoPj. When P lies at Pq beyond P 2 , the two parts are P\Po 
and PoPj. When P lies at P* beyond Pi, the two parts are 
P1P0 and P0P2. The parts are always read as here, that is, 
from the initial point to the division point and from the 
division point to the terminal point. 

When the line segment P1P2 is divided internally, both 
P1P0 and P0P2 are read in the same direction, and therefore 

P P 
the ratio p p is positive, and has a small value when P is 
* 0* 2 

near Pi, and a large value when P is near P 2 , that is, the 
value of the ratio is between and + » . 

When the line segment P1P2 is divided externally the two 
parts are read in opposite directions, and therefore the ratio 
P P f P P w 

WWf or BTjd°, is negative. Further, when the point of 

'0*2 *0* 2 

division lies beyond P% the ratio is between — » and — 1, 
and when the point of division lies beyond Pi the ratio is 
between — 1 and 0. 

It remains to express these geometric ideas analytically. 
This is done in the next articles. 

EXERCISES 

1. Upon a straight line locate two points Pi and P* 6 units apart. 

P P P P 

Locate a third point P such that ^^ = §. Such that —^ = — f . 

Suggestion. — These may be determined by methods of plane geometry, 
or may be computed by algebra. 

2. Divide a line 4 in. long into two parts that are in the ratio 3:1. 
In the ratio —5. 

21. To find the coordinates of a point that divides a line 
segment in a given ratio. 

Example 1. Internal paint — Required the coordinates of the point 
that divides the line segment from Pi( — 2, —4) to P*(5, 6) in the ratio |. 

Solution. — Draw a pair of axes as in Fig. 14, and locate the points 
Pi and Pi. Let Po(x , y ) be the required point. Draw lines through 
these points parallel to the y-axis_ and cutting the x-axis in Afi, Mi, 
and Mo respectively. 



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§21] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 21 



Then, by plane geometry, ' . / 

Mom s 



PiPo 
P0P1 



p.(M) 




**X 



But M iM o — «o — ( — 2) and MoJlf » - 5 — So, by [2J, for the abscissas 
of Af i, Af , and Mt are respectively —2, a? , and 5. 

And it is given that p^ — 5. 

Honflfl a? - (-2) 5 

Hence 5-Xo = 2' 

Solving this equation, jr » 3. 

Similarly, draw lines parallel to the 
s-axis cutting the y-axis in Ni iVo, and 
Nt respectively. 

TKan AW PiP 

Then m; " PoK 

But N1N0 - yo - (-4) and ATotfi - 
6 - yo. [2,] 

Hence ^=-^4 ^ 

o — Vo * 

Solving this equation, y = 3-^. 

Therefore the point P has as coordinates (3, 3\). 

Example 2. External point. — Required the coordinates of the point 

that divides the line segment from Pi ( —3, 5) to Pa (2, —3) in the ratio — f . 

* Solution. — Locate Pi and P* as in Fig. 15. 

Since the ratio is — £, the point of 

division, P (zo, yo) must be farther from 

Pi than from P», and so is beyond P* as 

shown. 

Draw the lines PiAf i, P*Af *, and PoM 

as in example 1. 

T , M,Mo PiPo 5 

inen MoMt = PoP, * 3' 

But MiMo - xo - (-3) and M M t »' 

2-s . [2i]. 

„ s - (-3) 5 

Hence — = — = — «. 

2 — x 3 

Solving this equation, Xo = 9^. 



X'-H- 



t-H-X 




p o(*o»o) 



y 

Fig. 15. 

Then xr ^ 

But AT^o - yo - 5 and ATotfi - -3 - y . 

yo - 5 5 

3* 



Similarly, draw PiiV\, PjiVj, and PqN* 
Then Tr ^ r = p-p- - - g . 

[2d. 



Hence 



-3 - y 
Solving this equation, yo ■» — 15. 
Therefore the point P has as coordinates (9j, —15). 



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22 



ANALYTIC GEOMETRY 



[522 



X i i l i 




«Po(*o.»o 



Example 3. External point. — Required the coordinates of the 

point that divides the line segment from Pi (5, — 2) to P»( — 2, 4) in the 

ratio — #. 

Solution. — Locate Pi and P» as in Fig. 16. 

Since the ratio is — $, the point of division Pq(x , j/o) must lie nearer 

to Pi than to P», and so is beyond Pi as shown. 

Draw the lines PiMi, PjMj, and P Mo as in example 1. 

^_ M l Mo PiPo _2 

inen M M* = PoP, * 5* 

But M iAf o = x — 5 and 

MoJif i - -2 - a*. [2i]. 

xx 2o — 5 2 

Hence — s = —v. 

—2 — Xq 5 

Solving this equation, x — 9$. 

Similarly, draw PiiVi, PiiVj, and 

PoiVo. 

Then iMr; = PoP; = ~s- 

But NiNo - yo - (-2) and 
tfoAT, - 4 - y . [2,]. 

Hence *+* = 4 

Solving this equation, y — —6. 

Therefore the point P has as coordinates (9§, —6). 

22. Formulas for finding coordinates of point that divides 
a line segment in a given ratio. — Required the coordinates 
of the point that divides the line segment from Pi(xi, yi) to 
P2O&2, y 2 ) in the ratio ri:r 2 
Let Po(x y yo) be the required point. 

Draw lines through Pi, P 2 , and P parallel to the y-axis and 
intersecting the z-axis in M\> M 2 , and M respectively. 
M1M0 _ P1P0 _n 
r% 

[2J 



Then M0M2 ~~ P0P2 

But M 1M0 = x — xi and M<Mi = x* — x . 

Xo "" Xl f*i 

£2 ~~ x <> r i 

Solving for x 0} x = * 2 , — — • 
ri + r a 

Similarly, draw lines through Pi, P 2 , and P parallel to 



Hence 



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§22] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 23 

the x-axis and intersecting the y-axis in N h iV a , and 
iV respectively. 

tv^ N * N * _ PJP± _ ri 

inen iV~o# 2 " P0P2 " r,' 

But iViiVo = 2/0 — 2/1 and iWVi = 2/2-00. [2*] 

Hence *^» = *• 

2/2 - 2/0 r 2 

Solvingforyo, »» - r ? + l* 1 

ri •+■ r 2 

Therefore the coordinates of P are 
rAi ~ - ^1X2 + r 2 Xi riy 2 + rtyi 

ri + r 2 ri + r 2 

Special case. — It is frequently required to find the coordi- 
nates of the point bisecting a line segment. In this case the 

two parts are equal, and the ratio — = 1. Formula [4] 

then becomes 

[6] x, = *+*, yo =y±±i>. 

It is readily seen that the results of the last two articles are 
true for oblique axes as well as for rectangular axes. 

EXERCISES 
In the first four exercises draw the figure, and solve without using 
formulas [4] and [6]. 

1. Find the coordinates of the point which divides the line from 
(-5, -8) to (-1, 4) in the ratio 3 : 1. 

2. Find the coordinates of the point which divides the line from 
(-1, 4) to (8, 1) in the ratio 1:3. 

3. Find the coordinates of the point which bisects the line from (8, 6) 
to (-2/ -3). 

4. Find the coordinates of the point which divides the line from 
(-4, 8) to (2, 6) in the ratio -|. 

6. Do each of the first four exercises by the formulas. 

6. Find the coordinates of the point which divides the line from 
(3, -9) to (-1, 5) in the ratio 5:3. 

7. Find the coordinates of the point which divides the line from 
(-6, 8) to (3, -2) in the ratio 3:1. In the ratio -2:3. 



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24 ANALYTIC GEOMETRY [$23 

8. Draw a triangle the coordinates of whose vertices are (1, 1), (2, —3), 
and (—4, —6), and find the coordinates of the middle points of its sides. 

9. The coordinates of P are (2, 3) and of Q are (3, 4). Find the 
coordinates of R so that PR : RQ - 3 : 4. 

10. Draw the triangle with vertices at (3, 5), (—5, —3), and (9, —7). 
Find the lengths of its medians. 

11. Show that the line joining the middle points of two sides of the 
triangle, having as vertices the points (8, 6), (1, 1), and (4, —5), is equal 
to one-half the third side. 

12. Prove that the diagonals of the parallelogram whose vertices are 
(10, 4), (—3, 4), (-6, -6), and (7, -6) bisect each other. 

13. The middle point of a line is at (4, 6) and one extremity is at 
(—3,-2). Find the other extremity. 

14. Find the coordinates of the points that trisect the line from (2, 2) 
to (-7, -4). 

16. Show that the median of the trapezoid whose vertices are (0, 0), 
(a, 0), (6, c), and (d, c) equals one-half the sum of the parallel sides. 

ANGLES FORMED BY LINES 

23. The angle between two lines. — That one line forms 

an angle with another is a geo- 
metric idea, and does not neces- 
sarily depend upon whether or 
not the lines are considered as 
having a positive or negative 
sense, that is, direction. In 
order to express the facts analyti- 
cally, we start with the following : 
Definition. — The angle that a line h makes with a line l* 
is the angle, not greater than 180°, generated by revolving U 
in a positive direction until it coincides with h. 

In Fig. 17, both (a) and (fc), the angle <p is the angle that h makes 
with 1%. 

It follows that the angle that h makes with h is t — (p. 

The definition still holds when the lines do not intersect, 
that is, are not in the same plane, if it is understood that 1% 
revolves in a plane parallel to l u until it is parallel to h. 




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§24] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 25 



24. Inclination and slope of a line. — An important special 
case of the angle that one line makes with another is the 
angle that a line makes with the x-axis. This angle is called 
the inclination of the line. It is always measured from the 
positive part of the x-axis. 

Thus, in Fig. 18, a\ is the inclination of h v 
and a s the inclination of 1%. 



^ 




+-X 



Fio. 18. 



Definition. — The tangent of the 
inclination of a line is called the slope 
of the line. 

Thus, if to is the slope of a line and a its 
inclination, then m » tan a. 

Since the inclination may be any angle in the first or second 
quadrant, the slope of a line may have any real value either 
positive or negative, including and ± « . 

25. Analytical expression for slope of a line. 

Example. — Required the slope and inclination of a line I passing 
through the points Pi (2, 3) and P»(5, 6). 

Solution. — Locate the points Pi and P% 
and draw the line J as in Fig. 19. Through 
Pi draw a line parallel to the x-axis, and 
through Pt a line parallel to the y-axis. 
These lines meet at Q, and the angle QP\P% 
is the inclination of I. Then to = tan QP\P%. 
6-3 




But tan QPiP, - ^ 



5-2 
tan" 1 ! 



- 1. 
45°. 



Fig. 19. 



Hence to =» 1, and a 

It should be noted that, whatever 
the position of the points, the line 
drawn parallel to the x-axis is so drawn that an angle equal to 
the inclination is formed. 

26. Formula for finding the slope of a line through two 

points. — Required to find the slope of a line I in terms of the 

coordinates of two points P\(x\, yi) and P^x*, y 2 ) on the line. 

Let the line be in either of the positions shown in Fig. 20. 

In either case draw a line through Pi parallel to the rc-axis, 



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26 



ANALYTIC GEOMETRY 



[§27 



Pi<Vi)f r 



i<Vt) 



forming an angle equal to a as shown; and through P 2 draw 
a line parallel to the y-axis meeting the first line in Q. 
Then, whether the slope m of I is positive or negative, 

m = tan a = ~~rk = 

PiQ x 2 — xi 

If Pi and P* are inter- 
changed, the slope is — — — , 

X\ £j 

which equals Vt ~~ Vl - 

H X t - Xj 

Therefore, in any case, the 
formula is 

[6] m - tan a = iLZJl. 
Xi — x 2 

27. The tangent of [the angle that one line makes with another 
in terms of their slopes. — Required the tangent of the angle that 
line It, having a slope of m h makes with 1%, having a slope of m*. 

Let the inclinations of h and h be ct\ and a* respectively. 




Pi(*i.»> 



t4 




\+X 



I 



Fio. 20. 




+-X 



Fig. 21. 




*-X 



Then tan a\ = m h and tan a* = m*. 

There are two cases: case I when ai>a 2 , Fig. 21 (a); and 
case II when ai<a%, Fig. 21 (6). 

In each case, let <p be the angle that U makes with 1%.' 

Then, in case I, a\ = as + ^>, or <p = ai — as. 

And, in case II, a 2 = a\ + (180° — ^>), 
or <p = 180° + (ai - a*). 



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§28] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 27 

In either case, 

t . tan a x - tan a, m x - m 2 

tan ^ = tan (ai — a*) = — 



[7] tan =*p 



1 + tan ai tan a 2 1 + mim 2 
nil — mi 



1 + mim* 



28. Parallel and perpendicular lines. — // two lines are 
parallel, their slopes are equal, and conversely. 

If two lines are perpendicular to each other, the slope of one 
is the negative reciprocal of the slope of the other, and conversely. 

If line* h is parallel to line U, then ai = at, and mi = m%. 

Conversely. If m, = tn t , ?\~ m * = 0. 

Then tan <p = 0, and hence <p = 0. 
Therefore h and 1% are parallel by Art 23, 
If ii and U are perpendicular to each other, ai = a%+ 90°, or 
ai = a t - 90°. 

Then tan ai = tan (at + 90°) = — cot at = — t 

tan at 

Therefore m\ = , and m 2 = 

f?l 2 Wtl 

Conversely. If mi = , tan ai = — =* — cot at. 

* m 2 tan a 2 

But cot at = tan (90° - a 2 ) = -tan (a 2 - 90°), or, 
cot a 2 = -tan (90° + a*). 

Then tan ai = tan (a 2 - 90°), or tan ai = tan (90° + a*). 

From this ai = a 2 - 90°, or ai = 90° + a 2 . 

Hence either a 2 — ai = 90° or aj — a 2 = 90°. 

Therefore <p = 90°, and Zi and 1% are perpendicular to each 
other. 

The following are the important facts to remember: 
[8] For parallel lines, mi = m 2 . 

[9] For perpendicular lines, mi = , and m 2 = 

m 2 mi 

Example. — Find the angle that the line through (4, 5) and (—2, —4) 
makes with the line through (0, 4) and (—6, —8). 



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*p 



1 +1-2 
- (-}) = 



tan" 1 (-4) - 172° 52.4'. 



28 ANALYTIC GEOMETRY [§28 

Solution. — The slope of 1 1 is mi — t~t~o " 5* 

4+8 
The slope of /» is m% - r T — 2. 
o •+• o 

1 — 2 1 

Substituting in [4], tan <p — i J. i.g ™ "~ S* 

r 2°l 

EXERCISES 

1. Find the slopes of the lines through the following pairs of points: 

(1) (-4, -4) and (4, 4). (4) (-V5, V5> and (\/2, V^). 

(2) (-4, 3) and (-3, 2). (5) (-a, 6) and (c, d). 

(3) (5, 0) and (6, VS). (6) (\/3, 2) and (\/2, 3). 

2. Find the inclination of each of the lines of exercise 1. 

3. Find the slope of a line that is perpendicular to the line through 
the points (3, 4) and (-2, -3). 

4. Show that the line through (4, 2) and (3, 7) is perpendicular to the 
line through (8, 1) and (13, 2). 

5. Find the value of y so that the line through (3, 7) and (4, y) shall 
be perpendicular to the line through (9, 10) and (6, 8). 

6. Prove by means of slopes that the three points (6, —3), (2, 3), and 
(—2, 9) are on the same straight line. 

7. Find the value of x so that the three points (x, 6), (2, 8), and (4, 7) 
shall be on the same straight line. 

8. Express by an equation the fact that a line passing through the 
points (4, 5) and Or, y) has a slope of f. 

9. A line passes through the point (—4, 6) and has a slope of — f. 
Find the abscissa of the point on the line whose ordinate is —3. 

10. Express by an equation the fact that a line passing through the 
point (—3, —6) is perpendicular to the line through the points (—2, 7) 
and (4, 6). 

11. Express by an equation the fact that a line passing through the 
point (7, 2) is parallel to the line through (—6, —2) and (4, —7). Find 
the point on this line whose abscissa is —3. 

12. Two lines l\ and U make tan~4 and tan _1 ( — !) respectively 
with the x-axis. Find the angle that l\ makes with U. 

13. Find the slope of the line that makes an angle of 47° with the line 
having a slope of 0.3674. 

Suggestion. — Substitute <p «■ 47° and m 8 =* 0.3674 in [7] and solve 
for mi. 

14. Find the angle that the line through the points (—3, 6) and 
(4, —2) makes with the line through the points (1, 1) and (—7, —7). 



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§29] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 29 

15. A line passes through the point (4, 5) and is parallel to the line 
through the points (3, -2) and (-2, 5). Find where the line cuts the 
y-axis. 

16. A line I makes an angle of 30° with the line through the points 
(2, 3) and (6, 7). Find the slope of I. 

17. Show that the lines joining the middle points of the sides of a 
quadrilateral whose vertices are the points (5, —4), (3, 6), (— 1, 4), and 
(—3, —2) taken in order, form a parallelogram. 

18. Prove by means of the slopes of the sides that the quadrilateral 
whose vertices are the points (4, 2), (2, 6), (6, 8), and (8, 4) is a rectangle. 

19. A point is equidistant from the points (—5, —2) and (2, —5), and 
the line joining the point to (4, 2) has a slope of — J. Find the codrdi- 
nates of the point. 

20. A line passes through the point (4, 5) and has a slope of 0.7236. 
Find the ordinate of the point on this line having as abscissa —2. 

21/ The vertices of a triangle are Pi(3, 4), P*(-4, 3), and Pi(-1, -4). 
Find the angle of the triangle at the vertex P t . 

POLAR COORDINATES 

29. Location of points in a plane by polar coordinates. — 
Thus far only the first method mentioned in article 13 for 



PlP.O) 





Fig. 22. 

locating points, has been used. The second method, that by 
polar coordinates, has advantages over the cartesian system 
in certain oases. This method will now be explained. 

In polar coordinates we locate a point in a plane by giving 
its distance and direction from a given fixed point in the 
plane. Thus, in Fig. 22, given the fixed point in the fixed 
directed line OX, then any point P in the plane may be located 
by stating its distance OP = p from 0, and the angle through 
which OX must turn to coincide with OP. 
. Definiti<ms^--The fixed point is called the pole or origins 



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30 



ANALYTIC GEOMETRY 



[§29 



the fixed line OX the initial line, or polar axis ; the line segment 
OP = p is called the radius vector of P; and the angle 6 the 
vectorial or directional angle of P. Together, p and $ are the 
polar coordinates of P, and are written (p, 6). 




P X {W) 



-**x 



P* (V30 ) 



Fig. 23. 



iP.C-i.-ViT) 




A(-a.30°) 



Fio. 24. 



In order to use both positive and negative numbers as 
coordinates of points, the usual conventions of trigonometry 
as to positive and negative angles of any size are accepted. 
It is also agreed that the radius vector is positive if measured 
from along the terminal side of the angle 0, and negative if 
measured in the opposite direction. 

Thus, Pi (5,60°) is located as shown in Fig. 23, the angle being mea- 
sured counter-clockwise and the radius vector along the terminal side in 
the positive direction. 

To plot the point P*(4, —30°), the angle is measured clockwise and 
the radius vector positive. (Fig. 23.) 

The following points are plotted as shown in Fig. 24: Pi (—3, 30°) 
andP,(-4, -W. 



» (-5,240°) 



>(6,-300 W ) 




*-x 



Fig. 25 



From the above illustrations it is clear that one pair of polar 
coordinates determine one, and but one, point in the plane. 



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§29] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 31 

On the other hand, for a single point there are an indefinite 
number of pairs of polar coordinates. 

Thus, if only values of 6 numerically less than 360° are taken, then 
the four pairs of coordinates (5, 60°), (-5, 240°), (-5, -120°), and 
(5, —300°) all determine the same point as shown in Fig. 25. 

For convenience in plotting, polar coordinate paper ruled 
with concentric circles and radial lines, as shown in Fig. 26, 
can be obtained. The following points are shown plotted in 

105° 90° »co 



m 



Fio. 26. 

Fig. 26: P(5, 20°), Q(-6, 80°), fi(8, fcr), S(-7, fir), and 
T(-8, -fr). 

EXERCISES 

1. Plot the following points in polar coordinates: 

(1) (3,30°). (6) (3, -**). (11) (3,1). 

(2) (7, 120°). (7) (-4,*). (12) (-4, -2). 

(3) (-2,40°). (8) (-2, -»). (13) (5, -3). 

(4) (-6, 150°). (9) (2, 0). (14) (-6, -5). 

(5) (4, -75°). (10) (-6 fr). (15) Or, -») 

2. Give three other pairs of coordinates in which is numerically 
less than 360° for each of the following points: (1) (7, 30°), (2) (-3, |x). 



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32 



ANALYTIC GEOMETRY 



[§30 



3. The side of a square is 4 in. and the diagonal is taken as the polar 
axis with the pole at a vertex. Find the polar coordinates of the vertices. 

4. What is the locus of all points for which p — 5? For which $ = $*-? 
For which - Jir? 

30. Relations between rectangular and polar coordi- 
nates. — Let X'X and Y'Y, Fig. 27, be a set of rectangular 
coordinate axes; and let the polar axis OX be taken on the 
positive part of the s-axis with the pole at the origin. 

Let P be any point in the plane. Draw OP y and QP per- 
pendicular to X'X. Then by the definitions already given, 
OQ = x, QP = y, OP = p, and L XOP = 6. 

.r By trigonometry and geometry 

it follows that 







* 


i 




P 


x ' 






V 


V 




V 

. \ 


r 

9 




r' 


x < 


I 



[10] 



x = o cos 0, 
y = p sin 0, 
x* + y* = p 2 . 



By means of these formulas 
polar coordinates can be ex- 
pressed in rectangular codrdi- 
Fl °- 27 ' nates. 

Also by trigonometry and geometry it follows that 

[11] q - V* 2 + y 2 , 

X 



6 = tan- 



By means of these formulas rectangular coordinates can be 
expressed in polar coordinates. 

EXERCISES 

1. The origin in rectangular coordinates coincides with the pole in 
the polar system, and the x-axis falls upon the polar axis. Find the 
rectangular coordinates of the following points: (6, !*■), (—2, !*■), 
(-5, M, (6, W, (3, fr), (8, ix), (2, *), (6, fr). 

2. Find the rectangular coordinates of the points whose polar coordi- 
nates are: (2, 40°), (3, 70°), (6.5, -30°), (1.2, 130°), (-4.5, 155°). 



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§31] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 33 

3. Find two pairs of polar coordinates for each of the following: 
(4, 4V3), (-3, -3), (3, 5), (-y% VS). 

4. By means of [10] derive d « Vpi* + pi* — 2pipi cos (0i — 0») from 
[3]. Here (xi, y\) and (pi, 0i) are the same point, as also are {x%, y t ) 
and (j>t, 0*). 

5. Derive the formula for the distance between two points in polar 
coordinates directly by means of the cosine theorem of trigonometry. 

6. The polar coordinates of P are (5, 75°) and of Q are (4, 15°). Find 
the distance PQ. 



TRANSFORMATION OF RECTANGULAR COORDINATES 

31. Changing from one system of axes to another. — From 
the discussions already given, it is evident that the coordinate 
axes may be chosen at pleasure. In any particular case it is 
clear that they should be so chosen that they can be used to 
the best advantage. In order to discuss certain problems that 
occur in analytic geometry, it is necessary to express the 
coordinates of points in the plane in another system of 
coordinates than that in whifeh they are already expressed. 

It is of advantage to deduce formulas for making these 
transformations which are of two kinds; 

(1) Transformation by trans- 
lation of axes, or changing to new 
axes that are parallel to the 
original axes. 

,(2) Transformation by rotation' 
of axes, or changing to new axes 
that make a certain angle with a 
the original axes. 

32. Translation of coordinate 
axes.— Let OX and OY, Fig. 28, 
be any system of cartesian axes; 

and let O'X' and O'Y' be another set parallel to the original. 
Let 0', the origin of the new system, have coordinates (h, k) 
when referred to the original system. 



N 


Y 


Y' 




i f«:*') 


B 


& 


1 


O 


A 


M ' 



Fio. 28. 



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34 



ANALYTIC GEOMETRY 



[§33 



Let P be any point in the plane, having coordinates (x, y) 
when referred to the original system and (a/, y') when referred 
to the new system. 

Draw a line through P parallel to the x-axis, intersecting OY 
in N and O'Y' in N'. Also draw a line through P parallel to 
the y-axis, intersecting OX in M and O'X' in M'. 

Then NP - 2W + tf'P, and AfP = MM' + M'P. Arts. 
10 and 18. 

But NP - x, iW = A, tf'P - x 7 , MP - y, MM' = fc, 
M'P = y'. 

Therefore the formulas for translating the axes are: 

[12] x = if + h, 

y = y' + k. • 

Solving these formulas for x' and y', 

[12 J * - x - h, 

y' - y - k. 

In this article it is not implied that the axes are rectangular, 
and therefore the formulas hold for transforming from any set 
of cartesian coordinate axes to a parallel set. 
33. Rotation of axes. Transformation to axes making 

an angle <p with the original.— Let 
imW) OX and Y, Fig. 29, be any system 
rX f of rectangular axes, and let OX' 
and OY 9 be another set of rec- 
tangular axes having the same origin 
as the original, but making an angle 
<p with OX and OY respectively. 

Let P be any point in the plane, 
having coordinates (x, y) when re- 
ferred to the original system, and 
(a/, y') when referred to the new system of axes. 

Join to P, draw MP perpendicular to OX, draw M'P 
perpendicular to OX'. Let LXOP = 6, and /.X'OP = 0', 
and OP = p. 




+>x 



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§33] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 35 

Then x = p cos 6 = p cos (0' + <p) 

= p cos $' cos <p — p sin 0' sin <p. [10] 
And y = p sin0 = p sin (0' + <p) = 

p sin 0' cos ^ + p cos 0' sin <p. [10] 
But x 7 = p cos 0' and y' ■» p sin 0'. [10] 

Substituting these values, 
[13] x = x' cos tp — y' sin ^, 

y = x' sin ? + y* cos ^. 
Solving these formulas for x' and y', 
[13i] *' = x cos <f> + y sin ?, 

y' = y cos ^ — x sin <p. 

Example.— The point P has coordinates (\/2; 2-\/2) when refeixed to 
a certain system of rectangular axes. Find the coordinates of P when 
referred to a new set of rectangular axes having the same origin but 
making an angle of 45° with the original. ! 

Solution. — Here we are to find x' and y' when x, y, and <p are known, 
and so we use formulas [13 J. 

Substituting in these formulas, 

x' - V2 cos 45° + 2\/2 sin 45° 

- V2 X iV2 + 2V2 X i V2 - 1 + 2 - % 
I/' = 2\/2 cos 45° - V2 sin 45° 

- 2V2 X i\/2 - V2 X i\/2 =2-1-1. 
Check the values by a drawing. 

EXERCISES 

1. Find the coordinates of the following points when referred to axes 
parallel to the original and with origin at the point (3, 4): (7, 8), (4, 3), 
(0,0), (-2,6), (-7, -5), (6, -8). 

2. Find the coordinates of the following points when referred to 
axes having the same origin as the original, but making an angle of 45° 
with them: (2, 3), (-3, 4), (-5, -5), (7, -1). 

3. The coordinates of the vertices of a triangle are Pi (—3, —4), 
Pi(6, —2), and P*(2, 7). Find the coordinates of the vertices when 
referred to parallel axes with origin at Pi. Plot. 

4. The coordinates of a point Pi are (3, 2). Find the coordinates 
of the origin of a new set of axes parallel to the old so that the coordi- 
nates of Pi shall be (—4, —6) when referred to the new axes. 

5. The coordinates of the vertices of a triangle are Pi(0, 0), P»(2, 2\/3)» 



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36 



ANALYTIC GEOMETRY 



[§34 



and P|( — 2, 4). Find the angle through which the axes must be rotated 
so that Pj shall lie on the new x-axis. Find the coordinates of the 
vertices of the triangle referred to the new axes. 
6. Derive formulas [13i] from Fig. 29 without solving [IS]. • 

AREAS OP POLYGONS 

34. Area of a triangle in rectangular coordinates. — Let 
APiPjPa, Fig. 30, be any triangle having vertices Pi(xi,yi), 

P*(x 2 , y % ), and P 8 (x 8 , yz) re- 



Af«fclTi) 




ferred to the axes OX and OY. 
Translate the axes to a new 
system having as origin one 
p 8 (**¥,)vertex of the triangle, say Pi 
(x h yi), 

Then the codrdinates of P 2 
and P 8 referred to O'X' and 
O'Y' are P 2 (x 2 ', y 2 ') and 
Pi{x*, yz) respectively, where 
Xi' = x t - xi, yj = yi - yi, 



^x' 



*»x 



Xi = Xz 



zi> V* = y» - yi 



Fio. 30. 

by[12J. 
Let ZX'O'P* = 2 , and /.X'O'Pz = 8 . 
The area of APiPjP* = \PiP% times the altitude from P a 

to PiP 2 . 

Hence area APiPaPa = \PiP% X PiP 8 sin (0 8 - 2 ) 

— \P\P% X PiPs (sin $z cos 2 - cos $z sin 2 ) 

= KP1P2 cos 2 X PiP 8 sin0 8 -PiPs cos 8 X P1P2 sin 2 ). 

ButPiP 2 cos 2 = x 2 ' = x% — Xi, PiPs sin 0* = y 8 ' = y z — 1/1, 

P1P3 cos 8 = x 8 ' = x 8 — Xi, PiP 2 sin 2 = y 2 ' = y 2 — y x . 

Substituting these values and putting A for area of APiP^s, 

^ = MO** - *0 (2/3 - yi) - (x 8 - Xi) (yi - 2/1)]. 

Multiplying and arranging, 

[14] A = £(xijr« - xjyi + x^ s - x 8 y 2 + x 8 yi - xiy 8 ). 



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§34] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 37 



This may be written in the determinant form 

Xi 2/1 1 
A = \ x 2 2/2 1 , 
x»y z l 

by which it can be readily remembered. The formula can 
also be remembered and the computation carried out by the 
following: 

Rule. — First, write down in a line the abscissas of the 
vertices taken in counter-clockwise order, repeating the first 
abscissa at the end; and under the abscissas write the cor- 
responding ordinates. 
Thus, x\ Xi xz x\ 

2/i 2/2 2/8 2/i 

Second, multiply each abscissa by the ordinate of the 
following column, and add. This gives xty* + x*yz +x 8 2/i. 

Third, multiply each ordinate by the abscissa of the follow- 
ing column and add. This gives yyx* + y&i + yzXi. 

Fourth, subtract the third from the second and divide by 2. 
This gives the area as in formula [14]. 

It is evident that the expression 
\P x Pt X PiPz sin (0 3 - 0a) for the area 
is positive or negative according as sin 
(03 — 02) is positive or negative. In 
order then to have the area positive, 
sin (08 — 02) must be positive. Hence 
03 — 02 must be positive, and 03>02. 
That is, P1P2 is turned counter- 
clockwise to coincide with P1P3. This 
will be true only if, in passing around 
the triangle in the order the vertices are taken, the area is 
always at the left as shown in Fig. 31. That is, a point 
moving around the triangle must move counterclockwise. 
Otherwise the area will be negative. 

Thus, in Fig. 31, the area of the triangle, if the vertices are taken in 
the order Pi, P%, P it is 

A - i[3(-5) - 9(-2) + 9-4 - 10(-5) + 10(-2) - 3-4] = 28i. 




P,(io.O 



+~x 



P,(9r5) 



Fig. 31. 



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38 



ANALYTIC GEOMETRY 



[§35 




If, however, the vertices are taken in the order P lt P%, P», the area is 
A - J[3-4 - 10(-2) + 10(-5) - 9-4 + 9(-2) - 3(-5)] = -28*. 

35. Area of any polygon. — Any polygon having its verticfes 
given in rectangular coordinates can 
be divided into triangles by diagonals 
drawn from any vertex. Its area can 
then be found. It can be readily 
shown that the area of any polygon 
can be found by the rule given for 
finding the area of a triangle. 

Thus, a polygon, Fig. 32, having 
five vertices as follows, ,taken in order 
counter-clockwise; Pi(xi,yi) 9 P%(xt 9 2/2), 
P*(xz, yz) Pa(x4, 2/4), and P 6 (x 6y 2/5), has its v area represented 
by the scheme, 

X\ x% x$ x& av#i 

y\ y% y* y\ 9*fo 

which evaluated by the rule gives 

-A = |[(xi2/2 + xjyi'+ x z y 4 + x 4 y b + x 5 2/i) 

- (2/1X2 + 2/2X3 + 2/3X4 + 2/4X5 + 2/5X1)]. 
Example. — Find the area of the polygon 
having the following vertices: (—3, 6), 
(2,-4), (8,1), and (4, 7). 

Solution. — Arranging the abscissas and 
ordinates, 

-3 2^4 
6-417 

A - §U(-3)(-4) + 2 1 + 8-7 + 4-6] - 
[6-2 + ( -4)8 + 1-4 + 7( -3)1 } = 65* square 
units. 

EXERCISES 
1 
in each case draw the figure: 

(1) (0,0), (10, 12), (-6,8). 

(2) (-4,6), (-2,9), (10, -4). 

(3) (17,2), (-3,9), (-6, -10) 

(4) (0,7), (10, -3), (-4,9). 



-3 
6 



Fig. 33. 
Find the area of the triangles having the following points as vertices, 




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§36] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 39 

2. Find the area of the quadrilateral with vertices (2, 5), (—7, 9), 
(-10, -3), and (6, -9). 

3. Find the area of the pentagon with vertices (1, — 2), (3, 1),(6, 2), 
(4, -4), and (2, -5). 

4. Show that the area of the triangle with vertices (2, 4), (— 6, — 8), * 
and (1#, —4) is four times the area of the triangle formed by joining the 
middle points of the sides. 

5. Find the area of the isosceles triangle with vertices (4, 5), (10, 13), 
and (4, 15). Find the altitude from the vertex at (4, 5), and find 
the' area as one-half the product of the base and altitude. Do the two 
results agree? 

6. Find the area of the triangle with vertices Pi(7, 9), P»(— 6, -8), 
and Pi (4, —6). Find the point P 4 dividing PjPj in the ratio 2:3, and 
show by areas that the triangle is divided into two triangles the areas 
•f which are in the ratio 2:3. 

7. If the vertices of a triangle in polar coordinates are Pi (pi, 0i), 
Pj(ps, #*), and P t (p if # t ), derive a formula for its area. 

Suggestion. — In [14] put Xi = pi cos $ h y x — p x sin e lf and similarly 
for P s and P s . Arrange and apply the subtraction formula for sines. 

8. Find the area of a triangle the vertices of which in polar coordinates 
are (10, 30°), (-12, 120°), and (6, 135°). 

APPLICATIONS 

36. Analytic methods applied to the proofs of geometric 
theorems. — One of the necessary conditions for the mastery 
of a mathematical subject is a thorough understanding of 
the fundamental ideas and methods. In the present chapter, 
stress has been laid upon the expressing of geometric ideas 
in an analytic form. Time will be well spent in reviewing 
these methods until they are fully comprehended. 

As will be repeatedly found in Subsequent chapters, analytic 
geometry gives a powerful method for treating a great variety 
of geometric questions. As an illustration of this a few 
elementary examples of the application of algebra to geometry 
are given in this article. 

One of the great advantages of the analytic method of 
solving geometric problems lies in the fact that an analytic 
result obtained by the simplest arrangement of the axes with 



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40 



ANALYTIC GEOMETRY 



[536 



Pt(*t.*t) 



reference to the geometric figure holds equally well for all other 
arrangements of the axes. 

It is well then always to make use of the simplest relations 
between the geometric figure and the coordinate axes. 

Example 1. — Prove analytically that the 
line segment joining the middle points of 
two sides of a triangle is equal to one-half 
the third side and is parallel to it. 

Given any triangle OPiPt, and AB bisect- 
ing OP* and PiP,. 

To prove AB = \OP u and that AB is 
parallel to OPi. 

Proof. — Choose the coordinate axes with 
origin at and the 2-axis along OPi. Then 
the coordinates of are (0, 0), and Pi and 
Pi may be designated by (xi, 0) and (x», y») respectively. 

Codrdinatesof A andBare fe ^\ and fo + Z \ ^\ respectively. [5] 
length of AB - VH^ 4) '+(§-!"' = * 9] 




PxlmiJh* 



But 



OPx - xi - - Xi. 
. AB = iOPi. 



[6] 



P%t*%*4 



Also slope of AB = 0, and slope of OP x =* 0, 
.\ AB is parallel to OPi. 

To see the desirability of this choice of the axes, the student should 
write out the proof when the 
vertices of the triangle are (x h yi), 
(xi, y%), and (x h yi). 

Example 2. — Derive a formula 
for the center of gravity of a 
triangle with vertices Pi(xi, yi), 
Pt(xt, Vt), and Pi(x«, y*); it being 
known that the center of gravity 
of a triangle is at the intersection "" 
of its medians, which is two-thirds 
of the length of any median from 
a vertex. 

Solution. — Here no choice of axes can be made that will simplify the 
work. 




*•<•** 



Pll*h*l) 



Fig. 35. 



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§36] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 41 

Choose any median as P\Q. Then it is required to find the coordinates 
(*o, Vo) of P such that PiP : PoQ -2:1. 

Codrdinates of Q are £* + *, UL ^ Jl ) ' 

By[ 4] ^- ^ + ; +ap s «nd^- y '+;+^ 

EXERCISES 

1. Use the formulas derived in example 2 and find the codrdinates 
of the center of gravity of a triangle with vertices (2, 6), (-8, 3), 
and (-4, -3). 

Prove the theorems in exercises 2-12. 

2. The diagonals of a rectangle are equal. 

3. The diagonals of a parallelogram bisect each other. 

4. The medians of a triangle intersect in a point which is two-thirds 
the length of any median from a vertex. 

6. The middle point of the hypotenuse of a right triangle is equally 
distant from the three vertices. 

6. The diagonals of a square are perpendicular to each other. 

7. If the diagonals of a parallelogram are equal, the figure is a 
rectangle. 

8. The distance between the middle points of the non-parallel sides 
of a trapezoid is equal to half the sum of the parallel sides. 

9. The lines joining the middle points of the successive sides of any 
quadrilateral form a parallelogram. 

10. The lines joining the middle points of the successive sides of any 
rectangle form a rhombus. 

11. In any quadrilateral, the lines joining the middle points of the 
opposite sides, and the line joining the middle points of the diagonals 
meet in a point and bisect each other. 

12. The sum of the squares of the four sides of a parallelogram is 
equal to the sum of the squares of its diagonals. 

13. Given Pi any point in the plane of a rectangle, prove that the 
sum of the squares of the distances from Pi to two opposite vertices 
of the rectangle is equal to the sum of the squares of the distances from 
Pi to the other two vertices. 

GENERAL EXERCISES 

1. If the points A, B, C, D, and E are any points on the same straight 
line, show that: 



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42 ANALYTIC GEOMETRY [§36 

(1) AB +CD -CB -ED - AE. 

(2) AE + EB + DE + EC - DB - ilC. 

(3) AC - J£B + CB - AE - 0. 

2. If the coordinates of the vertices of a rectangle are (0, 0), (8, 0), 
(8, 6), and (0, 6), what will be the oblique coordinates of its vertices if 
the ff-axis is the diagonal through the origin, the a>axis remaining as 
before? 

3. What are the oblique coordinates of the vertices of the rectangle 
of exercise 2 if the y-Qjda is taken as the diagonal through the point 
(0, 6)? 

4. What are the coordinates of the vertices of a square if a side is 
4\/2. and its diagonals are taken as the coordinate axes? 

6. A rhombus lies wholly in the first quadrant, and the angle between 
two of its sides is 30°. If the coordinates of two of its vertices are (0, 0) 
and (a, 0), find the coordinates of the remaining vertices. 

6. Find the coordinates of the vertices of an equilateral triangle of 
side a if its center is at the origin and the y-axis passes through one 
vertex. 

7. The angle between two oblique axes is 135°. Find the distance 
between the points (1, 3) and ( — 1, —3). 

8. What is the ratio in which the j/-axis divides the line segment 
joining (-2, 3) to (5, -1)? 

9. Find the coordinates of two points which divide the line segment 
from (2, 4) to (8, —8) internally and externally in the ratio whose 
numerical value is 2. 

10. find the coordinates of Pi and P* where Pi is on the positive 
j/-axis, Pi on the positive x-axis, and the point (2, 3) divides PiPj in 
the ratio 2:1. 

11. The point (-2, -2) divides the line PiPj in the ratio -4:3. If 
Pi has the codrdinates (2, 6), find the coordinates of P*. 

12. If Pi has the codrdinates (1, 4) and P s the codrdinates (5, 1), find 
a point P 8 on P1P2 such that PiPj will be a mean proportional between 
PiPs and 25. 

13. Prove analytically that the diagonals of a rhombus intersect at 
right angles. 

14. The hypotenuse of a right triangle is the line joining ( — 1, —2) 
to (6, 4). Find the coordinates of the third vertex if it lies on the x-axis. 

16. One end of the line whose length is 5 is at (4, 2). The abscissa 
of the other end is 7, what is its ordinate? 

16. The end points of a diagonal of a parallelogram are (2, —3) 
and (3, 2). Find the codrdinates of the remaining vertices if they are 
on the x-axis and y-axis respectively. Why is there only one solution? 



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§36] GEOMETRIC FACTS EXPRESSED ANALYTICALLY 43 

17. The codrdinates of the end points of one diagonal of a rhombus 
are (0, 0) and (2, 4). If one side lies along the positive x-axis, find the 
codrdinates of the end points of the other diagonal. 

18. Two points Pi and P% are at the same distance from the origin. 
If their polar codrdinates are (p, 0i) and (p, 0*), show that the slope of 

the line joining them is —cot — - — • 

19. What does the slope of the line joining ( — 1, 3) to (6, 7) become 
if the axes are rotated through an angle ? » tan" 1 }? 

20. What does the slope of the line joining (4, 3) to (—5, 6) become 
if the axes are rotated through 30°? 

21. What is the slope of the line through the points the polar 
codrdinates of which are (6, 30°) and (4, 60°)? 

22. Find the area of a triangle the polar codrdinates of the vertices oft 
which are (J*-, Jx), (*•, i*-), and (2r, i*-). 

23. Find the area of a triangle the polar codrdinates of whose vertices 
are (1, 60°), (3, 210°), and (2, 240°). 

24. A rectangle of sides 5 and 12 lies entirely in the second quadrant, 
with one vertex at the origin and the longest side on the negative z-axis. 
Find the codrdinates of its vertices if the axes are revolved so that the 
y-axis coincides with one diagonal. 

25. The codrdinates of the vertices of a parallelogram are (0, 0), 
(4, —3), (5, 0), and (1, 3). What will be the coordinates of its vertices 
if the axes are rotated so that the a>axis coincides with the longest side? 



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CHAPTER III 
LOCI AND EQUATIONS 

37. General statement. — In the present chapter will be 
considered some of the more simple cases of the first two funda- 
mental questions of analytic geometry as stated in article 3. 
The locus of an equation will be considered first, and then the 
equation of a locus. That is, the geometric interpretation of 
an equation will be dealt with first. 

38. Constants and variables. — A constant is a number that 
never changes, or one that does not change in the course of a 
discussion. 

Constants that never change are definite numbers, as 2, § , 
\/2, log 3, and x. Numbers that are constant during a dis- 
cussion, but may be different in another discussion are repre- 
sented by the letters that are assumed to have known values. 

A variable is a number whose value changes arbitrarily,* or 
according to some law. 

The number expressing the speed of a train as it gains 
headway is a variable. The price of a stock may change 
from day to day, and is expressed by a variable. The velocity 
of a falling body changes from instant to instant, and is 
expressed by a variable. 

If two variables are so related that for every value of one 
there is a corresponding value of the other, then the one is 
said to be a function of the other. 

Thus in the formula for the area of a circle, A = irr 2 , for 
every value of r there is a value of A. Then A is a function of 
r. This is written A = /(r). Likewise r may be considered a 
function of A. 

44 



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$39] ' LOCI AND EQUATIONS 45 

EXERCISES 

1. Give various illustrations of variables and constants. 

2. In the formula, A = $*r», for the volume of a sphere, which are 
constants and which variables? Is A a function of r? Is r a function 
of A? 

3. In the equation x + y =» 6, can either x or y be assigned values 
arbitrarily? Can both be given arbitrary values at the same time? 
Is x a function of y? Is y a function of x? 

39. The locus. — If the location of a point is determined by- 
certain stated conditions, then the locus of the point is the 
geometric figure such that, (1) all paints of the figure satisfy 
the given conditions, and (2) all points that satisfy the given 
conditions are in the figure. 

In proving that a certain figure is the required locus, it is 
sometimes more convenient, instead of (2), to prove that any 
point not in the figure does not satisfy the given conditions. 

The conditions determining a locus may be stated in the 
language of geometry, or may be stated by an equation. 

In the more simple cases the locus can be given immediately 
from the conditions stated. 

EXERCISES 

1. What is the locus of a point that is equally distant from two fixed 
points? 

2. What is the locus of a point in a plane and at a constant distance 
from a fixed point in that plane? 

3. What is the locus of a point equally distant from two intersecting 
straight lines and in the plane determined by those lines? 

4. What is the locus of a point equally distant from three fixed points 
and in the plane determined by the three points? 

5. In rectangular coordinates, what is the locus of a point whose 
abscissa is 0? Whose abscissa is 5? Whose ordinate is —6? 

6. What is the locus of a point whose coordinates satisfy the equation 
x = 4? Which satisfy the equation x = y? The equation x + y — 0? 

40. The locus of an equation. — If an equation is the analy- 
tic statement of geometric conditions, then it follows from the 
definition of a locus. Art 39, that the locus of an equation is 



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46 



ANALYTIC GEOMETRY 



[J41 



the locus of all points whose coordinates satisfy the equation; 
and, conversely, the coordinates of all points on the locus must 
satisfy the equation. 

While the preceding statement is general, only rectangular 
coordinates will be used in the present chapter. 

The drawing of the locus is spoken of as plotting the equa- 
tion! or plotting the locus of an equation. The locus is 
called the graph of the equation. 

41. Plotting an equation. — The steps in plotting an equation 
are: 

(1) Solve the equation for one, or each, variable in terms of the 
other. 

(2) Assign convenient values to 
one variable and determine corres- 
ponding values of the other vari- 
able, and arrange in a table. 
1 1 iMn 1 I f i » x (3) Choose a suitable unit and 
plot the pairs of values of the 
variables. 

Fig. 36. (4) Connect these points by a 

smooth curve. 
The variable to which values are assigned arbitrarily is 
called the independent variable. The other is then called the 
dependent variable. 
Example 1. — Plot the equation 2x + Zy — 13. 

(1) Solving for y, y = ^ 

(2) Assign values to x as shown in the following table and determine 
the corresponding values of y. 




z 


-4 


-2 


-1 





1 


2 


3 


5 


6i 


8 


10 


y 


7 


5f 


5 


4i 


31 


3 


2i 


1 





-1 


-2f 



(3) Locate a pair of rectangular coordinate axes, choose a suitable 
unit, and plot the points Pi(-4, 7), P 8 (-2, 5}), Pi(-1, 5), • • • 
Fig. 36 RA. as shown in Fig. 36. 



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§42] 



LOCI AND EQUATIONS 



47 



(4) Draw a smooth curve through the points. 

The curve is the locus of the equation and appears to be a straight 
line. 

Example 2. — Plot the locus of the equation x* + y % — 4x — 21. 
(1) Solving for y in terms of x and for x in terms of y, 



y - ± V21 + 4a; -a: «, and x - 2 ±V25 - y*. 

From the first it is readily seen that y is imaginary when x< —3 or 
when x>7, for then 21 + 4x — a?*<0 and the square root is imaginary. 

likewise, z is imaginary when y < — 5 or when y>5. 

It is evident then that we should not choose values of x less than 
— 3 nor greater than 7. And should not choose values of y less than 
—5 nor greater than 5. 

(2) Here it is convenient to assign arbitrarily some values to x and 
some values to y, in each case computing the corresponding values of the 
other variable. 



ip 

ill 



X 


y 


y 


X 





±4.6 


| 


7 or -3 


2 


±5 


±2 


6.6 or -2.6 


4 


±4.6 


: ±4 


5 or —1 


6 


±3 






-2 


±3 


; 





Fig. 37. 



(3)' The points are plotted as shown in Fig. 37. 

(4) A smooth curve is drawn connecting the points. This is the locus 
of the equation and appears to be a circle. 

42. The imaginary number in analytic geometry.— In the 
plan for plotting numbers in analytic geometry no method is 
provided for plotting imaginary numbers. It follows then 
that if one, or both, of the values of the variables satisfying 
an equation are imaginary or complex numbers, no point can 
be plotted having these as coordinates. Such numbers are 
often said to locate imaginary points on a curve. 



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48 ANALYTIC GEOMETRY [§43 

Some equations, such as x 2 + y 2 = 0, are satisfied by only 
one pair of real values for x and y. The locus of such an 
equation is a single point. 

Thus, x* + y* s= is satisfied only by x = and y = 0. 

Other equations, such as x 2 + y 2 + 4 = 0, are satisfied by 
no real values for x and y. The locus of such an equation is 
wholly imaginary. 

EXERCISES 

Plot the loci of the following equations: 

1; y = 2x + 4. 12. x* + 3y* = 0. 

2. 2x + 3y - 8. 13. s l + y* + 12 - 0. 

3. a; - 2y - 6 - 0. 14. x* + 2y* - 8. 

4. 3x - 4y - 5 - 0. 15. s l + y l + ftr - 7 

5. 16s - 3y - 42. 16. y» = 4* 1 . 

6. x* + y* - 25. 17. 9s l - 4y* = 36. 

7. x* + y* - 18. 18. 9a* + 4y l - 36. 

8. y* = 4a:. 19. 4y l - 9a* - 36. 

9. y* - 4a; + 3. 2Q. y = a* 

10. x 2 - 8y. 21. y = a* + 3. 

11. a* - y* = 4. 22. y = a* - 2s* + 6a; - 3. 

23. Plot the following equations upon the same set of axes: 

(1) x* + y* - 16, (2) x* - y* - 16, (3) y* - x* = 16, (4) -x* - y* = 16. 

24. Plot the following equations upon the same set of axes: 
(l)x« = 2y, (2) x* - -2y, (3) y» = 2s, (4) y* - -2s. 

DISCUSSION OF EQUATION IN RECTANGULAR COORDINATES 

43. Geometric facts from the equation. — Since it is 
possible to plot but a few points of a curve, the method of 
determining the curve by points is sufficiently accurate only 
in the case of simple curves. In general, much help in learn- 
ing the properties of a curve is gained by a study, or discussion, 
of the equation. First, it gives exact information regarding 
the curve; second, it furnishes a test of the accuracy of the 
plotting; and, third, it usually lessens the labor of plotting. 



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§44] LOCI AND EQUATIONS 49 

The properties of the locus of an equation that can be 
studied to the best advantage by analytic geometry are the 
following: 

(1) The intercepts of the curve. 

(2) The symmetry of the curve. 

(3) The extent of the curve. 

Various other properties can be studied by methods of 
the calculus. 

44. Intercepts. — The x-intercepts of a curve are the abscis- 
sas of the points where the curve intersects, or meets, the 
x-axis. The y-intercepts are the ordinates of the points 
where the curve intersects, or meets, the y-axis. Together 
the ^intercepts and the ^-intercepts are called the intercepts 
of the curve. 

Evidently, the s-intercepts are- found by putting y = in 
the equation and solving for x. Likewise the y-intercepts are 
found by putting x = and solving for y. 

It follows that, if an equation contains no constant term, 
the curve passes through the origin. 

Example. — Find the intercepts for the equation 16x* -f 25y* = 400 
Putting y - 0, 16x* - 400, or x - ±5. 

Putting x = 0, 2by % = 400, or y = ±4. 

.*. the it-intercepts are +5 and —5, and the y-intercepts are +4 
and —4. 

' EXERCISES 

Find the intercepts for the following equations: 

1. 2x - Zy = 10. 5. 5x*y - 15s + 4y =» 0. 

2. x* + y* = 36. 6. y - — -^ ~-^ 



(*+2)(*-l) 

3. 4x* + y* = 64. 7. y* - (* + 2)(* - l)(x - 3). 

4. 4x* -f y* - 8s - 2y + 1 - 0. 8. xy - 6. 

46. Symmetry, geometrical properties. — Two points are 
said to be symmetrical with respect to a given point when the 
given point bisects the line joining the two points. The given 
point is called the center of symmetry. 

Two points are said to be symmetrical with respect to a 



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50 



ANALYTIC GEOMETRY 



[$46 



given line when the given line is the perpendicular bisector 
of the line joining the two points. The given line is called 
the axis of symmetry. 

Thus, in Fig. 38, if Q bisects PiP,, Pi and f s 

Pi are symmetrical with respect to Q. Also, 

if A £ is the perpendicular bisector of PiPi, Pi a 

and Pi are symmetrical with respect to A B. 

If the points of a curve can be ar- *p f 

ranged in pairs which are symmetrical Fig. 38 

with respect to a line or point, then 
the curve itself is said to be symmetrical with respect to the 
line or point. 

Thus, in Fig. 39, the curve is symmetrical with respect to each of 
the coordinate axes and with respect to the origin. Tell why. 

EXERCISES 

1. Has a square a center 
of symmetry? Has a rec- 
tangle? A circle? A par- 

^allelogram? A regular 
hexagon? 

2. How many axes of 
symmetry has each of the 
figures of exercise 1? ^ 

3. In rectangular coor- 
dinates give the point that 
with each of the following is 

symmetrical with respect to the x-axis: (2, 4), (—2, 5), (—4, —2), 
(6, —8), (a;, y). With respect to the y-axis. With respect to the origin. 

46. Symmetry, algebraic properties.— In the preceding article 
symmetry has been considered from the side of geometry. 
It remains to determine how symmetry can be seen by an 
inspection of the equation. 

If a curve is symmetrical with respect to the z-axis, it 
follows that every point (x, y) on the curve has a correspond- 
ing point (x, —y) on the curve. Then the coordinates of the 




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§47] LOCI AND EQUATIONS 51 

point (x, — y) must satisfy the equation; that is, if — y is 
substituted for y, the equation reduces to the original form. 
It is evident that this occurs in an algebraic equation when 
only even powers of y appear in the equation. (See Art. 120.) 

Likewise the curve is symmetrical with respect to the 
y-axis if, when —x is substituted for x, the equation reduces 
to the original form. This occurs in an algebraic equation 
when only even powers of x appear in the equation. 

Since the pair of points (x, y) and (— x, —y) are symmetrical 
with respect to the origin, it follows that if, when —x is sub- 
stituted for x and — y for y, the equation reduces to its original 
form, the curve is symmetrical with respect to the origin. 
It is evident that this occurs in an algebraic equation if each 
term is of even degree, or if each term is of odd degree, in x and 
y. In applying this test a constant terin is considered as of 
even degree. 

It also follows that if a curve is symmetrical to both coordi- 
nate axes it is symmetrical with respect to the origin. 

EXERCISES \ 

State for which of the following equations the curves are symmetrical 
with respect to the z-axis, the y-axis, and the origin. \ 

1. Zx + y + 6 - 0. 7. x* + y = 6. 

2. x* + y* = 25. 8. x* + 2xy + y* = 9. 

3. 3s* - \y % m 12. 9. y* = (* + 1)(* - 2). 

4. x* + y* + 2x = 16. 10. xV + 4x 4 = 16. 

5. y» = 4x. 11. Z* + 4z + 2y + 3 - 0. 

6. x*y* = 16. 12. x* - x = y. 

47. Extent. — Under this heading we endeavor to find how 
the curve lies with reference to the coordinate axes by finding, 
first, for what values of either variable there are no points on 
the curve; and, second, for what values of either variable the 
curve extends to infinity. 

To do this the equation is solved for each variable in terms 



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52 



ANALYTIC GEOMETRY 



[§47 



of the other. First, if a radical of even index involves a 
variable, certain values of that variable may give imaginary- 
values for the other variable, in which case there are no points 
on the curve. If no radicals of even index are involved, 
there will be at least, one real value of either variable for a 
real value of the other. In which case there are points on 

the curve for every value of 
either variable. 

Second, if the solution for 
either variable gives rise to a 
fraction having the other vari- 
able in the denominator, then 
certain finite values of the second 
variable may make the first in- 
finite. If no such fraction occurs 
both variables may become in- 
finite at the same time. 




Example 1. — Investigate 9s* + 4y* 

=« 36 as to extent. 

Solving for x, x = + js /9 — y* . 
Solving for y, y = ± \ y/A - x*. 
Therefore x is imaginary when 
9 — y*<0, that is, when y<— 3, or 
when y > -f 3. 

And y is imaginary when 4 — x 2 <0, 
that is, when x< — 2, or when x> +2. 
The curve is then confined to the 
portion of the plane in which the 
abscissas do not exceed 2 in numerical value, and the ordinates do not 
exceed 3 in numerical value. 

Example 2. — Discuss the equation x* — ax — y — and plot the 
curve. 

Discussion. Intercepts. — Let x = then y = 0. 
Let y — then x* — ax =» 0. Solving this for x, x — or ± \/a . 
Hence the ^-intercept is 0, and the x-intercepts are and ± \/o • 
Symmetry.-- -Since all terms are of odd degree in x and y, the curve 
is symmetrical with respect to the origin. 
Extent. — Solving for y % y — .«• — ax. 



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§48] 



LOCI AND EQUATIONS 



53 



The letter a represents an arbitrary constant and may have any 
value assigned to it. But, in assigning a value, do not choose one that 
would cause a term to disappear. For the purposes of this discussion 
it is given the value 4. 

Since no even root is involved, either variable has a real value for 
any value of the other. 

Since large positive values of x make x 3 — ax large and positive, for 
such values of x, y increases as x increases. 

Likewise, for numerically large negative values of x, y decreases as 
x decreases. 

Plotting. — Tabulating coordinates for positive values of x, the curve 
can be located in the first and fourths quadrants and, by symmetry, 
in the second and third quadrants, and is as shown in Fig. 40. The 
arbitrary value assigned to a is 4, and the unit on the y-axis is one-fifth 
of that on the x-axis. 



X 





i 


l 


U 


2 


3 


4 


y 





-ii 


-3 


-2| 





15 


48 



EXERCISES 

Discuss each of the following equations and plot their curves. 



1. ** + y* = 64. 


10. xy + 12 = 0. 


2. x* - y* - 64. 


11. y = X s - 9x. 


3. 4x» + 9y» - 36. 


12. y(x* + 1) - 8 - 0. 


4. 4x* - 9y* - 36. 


13. 9x* - y*. 


5. y* = &x. 


14. y* - (x - 2)(* + 1)(* + 3) 


6. x* - Sy. 


15. y* = (x - l)*(x - 2). 


7. x 2 = Sy - 6. 


16. y{x - 1) - 1. 


8. re 2 + y* - 4* - 20 = 0. 


17. y* = ax 3 -f x 1 . 


9. xy = 15. 


18. x(x - 2a)* - ay* = 0. 



48. Composite loci. — Any function of the two variables 
x and y may be denoted by f(x, y). Then f(x, y) = is a 
compact way of writing any equation in these two variables. 

Theorem. — If the expression f(x, y) can be factored into 
variable factors, the locus of f(x, y) = consists of as many 
distinct curves as there are variable factors off(x f y). 

Proof. — Suppose /(x, y) can be factored into/i(x, y), /*(#, y), 
M*> V), ' ' ' • 



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54 



ANALYTIC GEOMETRY 



IH9 



Then AC*, »)/,(*, y)/,(«, y) • • - =0. 
Now any values of x and y that will make any one of these 
factors equal zero will satisfy the original equation. 
Hence all points on the separate loci of 

M*,v) - 0,/i(«, y) - 0,/,(z, y) - 0, • • • 

will also be points on the locus of 

/(*> y) - o. 

Much time is often saved when 

/(#, y) can be factored, by plotting 

each of the equations fi(x, y) = 0, 

/*(*, y) - 0, /,(», y) - 0, • • • 

separately. 




+~x 



Fig. 41. 



Example, — Plot the locus of the equation 
a* + xy* - 2x* - 2y l - 16x + 32 = 0. 
The factors of x* + xy* - 2s 1 - 2y* — 16s + 32 are 

(* - 2)(x* +y» - 16). 
Equating each factor to zero, x — 2 and a* + y* = 16. 
The first is a straight line and the second a circle as shown in Fig. 41 

EXERCISES 

Find a single equation whose locus is the combination of the loci 
of the separate equations in each of the following and plot. 

1. xy - 6 = 0, xy + 6 = 0. 

2. x - 2y + 3 = 0, x - 2y - 3 = 0. 

3. x = 0, x = 3, x = 5. 

4. x - y y x * + y* - 16. 

5. jc* -+- y* = 4, sy = 6. 

Plot the locus of each of the following by first factoring /(re, y), and 
then plotting each factor equated to zero. 

6. a5*y* - 16. 8. x 1 + 2xy + y* - 4 - 0. 

7. x* - y* - 0. 9. x* - 6s* + 11* - 6 * 0. 

10. x* + x*y - 4x - 4y + xy* + y» = 0. 

11. Plot the locus of (x* — x — 6)(y* + 2y — 8) =0, and show that 
the lines enclose a rectangle. 

49. Intersection of two curves. — The curves of two equa- 
tions are, in general, distinct, and may or may not intersect. 



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§50] LOCI AND EQUATIONS 55 

It follows from the definition of the locus of an equation that, 
if a pair of values satisfy both equations, they are the coordi- 
nates of a point of intersection. And, conversely, if the 
curves intersect, the coordinates of a point of intersection 
must satisfy both equations. 

In order then to find the coordinates of the points of inter- 
section, it is only necessary to solve the equations simultane- 
ously. Or, in order to find values of x and y that satisfy the 
equations simultaneously, the equations may be plotted and 
the codrdinates of the points of intersection determined from 
the figure. This is useful when the equations are such as 
cannot be solved simultaneously. 

EXERCISES 

Find the points of. intersection of the curves of the following pairs of 
equations by solving the equations. Check the results by plotting. 

1. x 1 + y* - 16, x + y » 0. 

2. x* + y* - 16, x* - y* = 9. 

3. x* - 4y l + 7 = 0, 2x + 3y - 12 = 0. 

4. x* + y l - 25, 9s* + 49y l = 441. 

5. x* + y l - 25, 27y* - 16rr*. 

6. Find the distance between the points of intersection of 

x* + y 2 * 12, and y* — 4x. 

7. Solve the following equations by plotting to find the codrdinates 
of the points of intersection: x 1 + y = 7, x + y % =* 11. 

EQUATIONS OF LOCI 

60. So far in the present chapter the problem considered 
has been the finding of the locus when the equation was given. 
Here the second fundamental question is taken up, that of 
finding the equation of a locus when the locus is known. 
That is, the algebraic statement is to be found when the 
geometric figure or description is known. 

Definition. — The equation of a locus is an equation such 
that (1) the codrdinates of every point on the locus satisfy the 
equation, and (2) every pair of values which satisfy the equation 
are the codrdinates of a point on the locus. 



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56 



ANALYTIC GEOMETRY 



[§51 



61. Derivation of the equation of a locus. — The process of 
deriving the equation of a locus depends largely upon the 
ingenuity of the individual. The following suggestions, how- 
ever, will be helpful, but it is not intended that it is necessary 
always to take these steps in order. 

(1) From the description of the locus sketch a figure involving 
ail the data. 

(2) Draw a pair of codrdinaie axes and select P(x, y) any 
point on the locus. 

Frequently the coordinate axes are determined by the data; 
but if they are not, they should be located so as to make the 
equation as simple as possible. 

(3) Write an equation between geometric magnitudes, using 
the conditions of the problem. 

(4) Express the geometric magnitudes of (his equation in terms 
of the codrdinates of P and the given constants, and simplify the 
resulting equation. 

The final equation will, in general, contain the variables 
x and y, and all the constants involved. 

(5) Show that any point whose codrdinates satisfy the equation, 
is on the locus, and thus show that the second requirement of the 
definition is fulfilled. 

A discussion of the equation will 
often give further facts concerning 
the locus. 

Example 1. — The locus of a point is a 
straight line passing through Pi (—2, 3) and 
having an inclination of 60°. Find its 
equation. 

Solution. — (1) Here the coordinate axes 
are determined by the data. In Fig. 42, OX 
and OF are the axes and PiP the locus. 

(2) P(x t y) is any point on the locus. 

(3) Slope PiP = tan 60°. 




»»x 



Fig. 42. 



(4) Rope PS -j^-? by [6]. 



■ F-3 

' 'x +2 
Simplifying, Zx - y/Zy + 6 + Z\/Z 



VI. 



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§51] 



LOCI AND EQUATIONS 



57 



(5) Any point P(x, y) whose coordinates satisfy the equation 
3x - y/Zy + 6 + 3 y/Z - 0, 
must also satisfy the preceding equation since this equation can be 
reduced to that form by reversing the steps. But the equation 



2,-3 



V3 



+~X 



x-f2 

simply says that the slope of a straight line through (x, y) and (—2, 3) 
is equal to y/Z> and hence its inclination is 60°. 

Therefore the equation of the locus is 

Zx - y/Zy + 6 + 3\/S = 0. 

Example 2. — Find the equation of the 
locus of a point that moves at a dis- 
tance 8 from the point (3, —5) and re- 
mains in the plane of the coordinate axes. 

Solution. — (1) In Fig. 43, the codrdi- p , x y ^ 
nate axes are drawn and the data located. 

(2) P(x, y) is any point on the locus. 

(3) By the conditions of the problem, 
PC = 8. 

But PC = y/{x - 3)* + (y + 5)» by 
[3]. 

(4) .*. V(x - 3)» + (y + 5) 2 - 8. 
Squaring, z J - 6x + 9 + y % + lOy + 25 - 64. 
Simplifying, x 1 + y* - 6x -f 10y - 30 - 0. 

This is the equation that is satisfied by the coordinates of any point 
on the locus. The proof of the converse is left to the student. 




EXERCISES 

Give orally the equations of the loci described in exercises 1 — 10. 

1. A point moves parallel to the y-axis and 4 units to the right. 
Parallel to the y-axis and 6 units to the left. 

2. A point moves parallel to the z-axis and 7 units above. Parallel 
to the x-axis and 3 units below. ' 

3. A point moves parallel to the z-axis and 3 units above the point 
(3, 6). Parallel to the a>axis and through the point (—6, 4). Parallel 
to the a>-axis and through the point (0, —7). 

4. A point moves parallel to the line y - 4 and 6 units above it. 

5. A point moves parallel to the line x = — 3 and 8 units to the right 
of it. 



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58 ANALYTIC GEOMETRY [§51 

6. A point moves so as to bisect the angle the y-axis makes with the 
x-axis. 

7. A point moves so as to bisect the angle the x-axis makes with the 
y-axis. 

8. A point moves so as to keep 6 units from the origin. 

9. A point moves so as to keep 8 units from the point (2, —1). 

10. A point moves so as to keep equidistant from the lines y ■= 8 
and j/ =» —2. 

11. Find the equation of the locus of a point that is equidistant from 
the points (5, 4) and (—6, —2). 

12. Find the equation of the locus of a point that moves at a distance 
10 from the point (-6, -8). 

13. Find the equation of the circle having its center at the point 
(3, 4), and passing through the point (7, 7). 

14. Find the equation of the circle having the extremities of a diameter 
at the points (-4, -6) and (2, 2). 

15. Find the equation of the perpendicular bisector of the line joining 
the points (—4, -8) and (5, 2). 

16. Find the equations of the perpendicular bisectors of the sides 
of the triangle whose vertices are the points (0, 0), (8, 6), and (—4, 10). 

17. Find the equation of the locus of a point that moves so as to keep 
four times as far from the x-axis as from the y-axis. Plot. 

18. Find the equation of the locus of a point that moves so as to keep 
three times as far from the point (2, 3) as from the point (—6, 2). 

19. A point moves so that its ordinate always exceeds ) of its abscissa 
by 8. Find the equation of its locus and plot. 

20. A point moves so that the sum of its distances from the points 
(3, 0) and ( — 3, 0) is 8. Find the equation of its locus and plot. 

21. A point moves so that the difference of its distances from the 
points (3, 0) and (—3, 0) is 4. Find the equation of its locus and plot. 

22. A point moves so that the difference of the squares of its distances 
from the points (—3, —1) and (—2, —4) is 5. Find the equation of the 
locus and plot. 

23. A point moves so that the slope of the line joining it to the point 
(—2, 3) equals twice the slope of the line joining it to the point (4, —2). 
Find the equation of the locus. 



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CHAPTER IV 



THE STRAIGHT LINE AND THE GENERAL EQUATION 
OF THE FIRST DEGREE 

62. Conditions determining a straight line. — In plane 
geometry it is found that two independent conditions deter- 
mine a straight line. Just so in analytic geometry any two 
conditions that fix the line will determine its equation. Since 
the same straight line can be determined in a number of 
different ways, it may be expected that there will be several 
forms of the equation for the same straight line. 

Some of the conditions that de- 
termine a straight line are the 
following: 

(1) A point on the line and the 
direction of the line. 

(2) Two points on the line. 

(3) The length and direction of 
the perpendicular from the origin to 
the line. 

Each set of these conditions gives rise to a standard form 
of the equation of a straight line. 

63. Point slope form of equation of the straight line. — 
Suppose the straight line I, Fig. 44, passes through the 
point P\(xi, yi), and that its direction is given by its slope 
m = tan a. If P(x 9 y) is any point on I, then the slope 
of PPi must be constant and equal to m. By [6], the 

slope m of PPi is m = V ~ Vl - 

X — X\ 

Clearing this equation of fractions, 

[16] y - yi = m(x - Xi). 

59 




Fio. 44. 



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60 



ANALYTIC GEOMETRY 



[§54 




*»x 



This is the point slope form of the equation of a straight line. 

Since P(x, y) is any point on I it follows that evejy point 
on I satisfies [16]. 

In order to prove that every point which satisfies [16] is 
on line I, let Pstes, yz) 9 Fig. 45, be such a point, then 

yz - yi = rn{x z - Xi). 

Dividing both sides of this equation by X* — Xi, 

y*- y\ 

= m. 

Xz — X\ 

This shows that the slope of the line PiP 8 = m. Therefore 
PiP 3 and I are parallel. Since PiPa 
and I pass through the same point Pi, 
the line P1P3 and I coincide. There- 
fore Pz lies on I. 

In the discussion of other forms of 
the equation of a straight line, the 
proof that every point whose coordi- 
nates satisfy the equation of the locus, 
is on the locus, is so similar to the proof just given that it will 
be omitted. Nevertheless this fact should not be lost sight 
of, for it is one of the essential conditions in determining the 
equation of a locus. 

64. Lines parallel to the axes. — In article 53 it is tacitly 
assumed that the line whose equation is to be found is not 
parallel to the y-axis. If it is, a equals 90°, m is infinite, and 
equation [16] is meaningless. If the line is parallel to the 
#-axis, it must cut the x-axis at some point (a, 0). Every 
point on this line has its abscissa equal to a, hence the equation 
of the line is 

x = a. 

Similarly every line parallel to the z-axis cuts the y-axis 
at some point, say (0, 6). Every point on this line has its 
ordinate equal to b and hence the equation of the line is 

y = b. 



Fio. 45. 



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§66] EQUATION OF THE FIRST DEGREE 61 

Example l.—Find the equation of a line through (—2, 3) and with 
an inclination of 136°. 

Substituting x\ — — 2, y\ — 3, and m = tan 136° — — 1 in [16], 
y-3 - <-l)<* + 2), 
or x + y — 1 - 0. 

Example 2. — Find the equation of a line through the point (2, 6) and 
parallel to the line joining the points (—3, 4) and (1, 5). 

By [6] the slope of the line joining the two points is J. 

Therefore the slope of the required line is also } 

Substituting m — }, Xi « 2, and y\ =» 6 in [16], 
the equation of the required line is y — 6 - \{x — 2), 
or x - 4y + 22 = 0. 

EXERCISES 

Find the equations of the lines determined by the following sets of 
conditions: 

1. Through (2, -3), slope J. 

2. Through (-2, -4), inclination 135°. 

3. Through (1, 5), inclination 120°. 

4. Through ( — 1, 2), parallel to the line 
joining (7, 6) to (2, 3). 

6. Through ( — 1, 2), perpendicular to the 
line joining (7, 6) to (2, 3). 

6. Through (3, 4), parallel to the t/-axis. 

7. Through (3, 4), parallel to the s-axis. Fio. 46. 

8. Through ( — 1, 2), inclination — tan -1 J. 

9. Through (1, —2), inclination = sin" 1 # . 

10. Through (3, 2), inclination = cos -1 ^. 

11. Find the equation of the tangent line to the curve y — «* — x, 
at the point whose abscissa is 2, if its slope equals 11. 

Suggestion. — Find the ordinate of the point whose abscissa is 2 and 
substitute in [16]. 

12. Find the equation of the tangent line to the curve y =■ 2x* — x + 3 
at the point whose abscissa is 2, if its slope equals 7. 

65. Slope intercept form. — In Fig. 46, let the intercept of 
the line on the #-axis equal 6 and let the slope of the line 
equal m. Since the y-intercept has the coordinates (0, b) 
this problem is a special case of the point slope form. 

Putting xi = and j/i = 6 in [15], then y — 6 = mx f or 
[16] y = mx + b. 




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62 ANALYTIC GEOMETRY [§56 

This is the slope intercept form of the equation of a 
straight line. 

66. Two point form. — Let the two points through which 

the line passes be P\(x\, y\) and Piixi, y*). Since Pi is a point 

on the line and m is the slope of P1P2, this form can be derived 

V\ — Vt 

from [16] by substituting for m its value — . Equation 

X\ %% 

[16] then becomes 

[17] y - yi - |^r < x " Xl >- 

Xi — Xj 

Note that this equation is not valid if X\ — x% = 0. The line 
is then parallel to the t/-axis and hence its equation is x = X\. 

This is the two point form of the equation of a straight line. 

Since the three points P, Pi, and Pi on the straight line 
through PiP 2 , always form a triangle whose area is zero, the 
equation of the straight line can be written in the deter- 
minant form by article 34, as follows: 



= 0. 



EXERCISES 

Find the equations of the lines given by the following sets of conditions: 

1. The y-intercept « 3 and the slope = J. 

2. The y-intercept = — 2 and the slope = 3. 

2 

3. The y-intercept = \ and the inclination = sin -1 — =• 

Vl3 

4. Passing through the points (1, 6) and (7, 2). 

5. Passing through the points (—2, 1) and (3, —4). 

6. Passing through the points ( — 1, —2) and (—4,-3). 

7. What is the effect on line [16] if b is changed while m remains 
unchanged? What is the effect if m is changed while b remains 
unchanged? 

67. Intercept form. — If the straight line cuts both axes, 
let its ^-intercept, Fig. 47, equal a and its ^-intercept equal 6. 

Its equation can be derived from [17] by replacing (xi, yd 
by (a, 0) and (x 2 , y 2 ) by (0, b). 



X 


y 


1 


Xi 


V\ 


1 


X2 


y* 


1 



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§57] 



EQUATION OF THE FIRST DEGREE 



63 



Equation [17] then becomes y = — (x — a). 

Multiplying both sides of this equation by j~t and transpos- 
ing the :c-term to the left hand side, it becomes 



[18] 



l+l-> 



This is the intercept form of the 
equation of a straight line. 

Care must be used in employing 
this form of the equation, since it is 
not valid if either or. both intercepts 
are zero. 

68. Normal form. — A line is completely determined if the 
length and direction of the perpendicular to it from the origin 
are known. 




Fig. 47. 




^^O 



Fig. 48. 

Let C, Fig. 48, be the foot of the perpendicular drawn to the 
line from the origin, and let (p, 6) be the polar coordinates of C 

Then OC = p and angle XOC = 0. 

Since the line AB is perpendicular to the line OC, its slope 
is the negative reciprocal of the slope of OC and equals —cot 0. 



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64 ANALYTIC GEOMETRY [§58 

Since the line AB passes through the point C, a point on AB 
is known. The rectangular coordinates of this point are 
(p cos 0, p sin 6). 

Hence the equation of AB can be found by substituting in 
[16], m = — cot 0, X\ = p cos 0, and y x = p sin 0. 

Making these substitutions, [16] becomes 

y — p sin 6 = —cot 6 (x — p cos 0). 

Multiplying both sides of the equation by sin and transpos- 
ing all terms to the left hand side, gives 

x cos + y sin — p(sin 2 + cos 2 0) = 0. 
[19] x cos e + y sin e - p « 0. 

This is the normal form of the equation of a straight line. 

If cot = » , the line is parallel to the #-axis and its equa- 
tion is x = p. But even in this case the normal form is valid, 
for if cot = » , = 0° and the normal form would read 
x cos 0° + y sin 0° — p = 0. Since cos 0° = 1 and sin 
0° = this equation is equivalent to x = p. 

Example. — Find the normal form of the equation of a straight line if 
$ » 30° and p » 6. 

Substituting these in [19], x cos 30° + y sin 30° - 6 - 0. 

Since the polar coordinates of C can be written either (6, 30°) or 
(—6, 210°), the normal form of the equation of this line could also be 
written x cos 210° + y sin 210° +6-0. That these equations are 
equivalent can readily be seen if the trigonometric functions are 
replaced by their numerical values. 

EXERCISES 
Find the equations of the following lines having given: 

1. a - 3, b - -2. 6. 6 - 60°, p « -3. 

2. a - -1, b - 6. 7. - 135°, p = 2. 
8. o - i 6 - |. 8. - 210°, p - 1. 

4. a - -i 6 - -i 9. - 330°, p - - 4. 

6. e - 60°, p - 3. 10. - 150°, p - - 2. 

11. What is the effect on line [19] if p is changed while remains un- 
changed? What is the effect if is changed while p remains unchanged? 



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§59] EQUATION OF THE FIRST DEGREE 65 

69. Linear equation. Theorem — Every equation of a straight 
line is of the first degree in one or two variables. 

Conversely. Every equation of the first degree in one or 
two variables is the equation of a straight line. 

Proof. — Every straight line intersects, does not intersect, or 
coincides with the y-axis. 

In the first case by means of article 55 its equation can be 
put in the form y = mx + b, in the second case its equation 
is x = a, and in the third case x = 0. Each of these equa- 
tions is of the first degree in x and y. 

Proof of converse. — Consider the most general equation of 
the first degree in two variables. This is 

[20] Ax + By + C - 0. 

Assume that B^O and solve this equation for y, it becomes 

Ax C 
* - - "F - F 

Comparing this equation with the form y = mx + b shows 
at once that it is the equation of a straight line whose 

slope m = — -^, and whose y-intercept 6 = — -g- 

Q 

If B = 0, [20] becomes Ax + C = 0, or x = - j This is 

the equation of a straight line parallel to the y-axis. 

Hence every equation of the first degree is the equation of a 
straight line. 

Ax + By + C = 0, the most general equation of the first 
degree in two variables, is called the general equation of a 
straight line. 

60. Plotting linear equations. — Since every equation of the 
first degree represents a straight line, it is sufficient in plotting 
the graph of such an equation to find two points which satisfy 
the equation and then join these points by a straight line. 
Usually two such points that can be easily found are the inter- 
cepts on the x and the y-axes. 

5 



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66 ANALYTIC GEOMETRY [§61 

61. Comparison of standard forms. — In article 59 it was 
seen how the general equation could be transformed into the 
slope intercept form. This method of transforming one 
form of an equation into another is of great use in analytic 
geometry, since by comparing the constants in two forms of 
an equation of a line or curve, much information can be 
secured. In this particular case the transformation from the 
general form to the slope intercept form, enables one to read 
off by inspection the slope and y-intercept of the line. 

For example, if the equation 3x + 4y = 12, be solved for y it becomes 
V - -fa; +3, 

Comparing this equation with y = mx +b f shows that the slope of 
the line is — f and its ^-intercept is 3. 

Take the same equation, Zx + 4y =» 12, and divide both sides of the 

equation by 12, and ^ + g ■■ 1. 

Comparing this equation with - + ? = 1 shows that the x-intercept 

is 4 and that the ^-intercept is 3. 

Here the a>intercept can be as easily found by putting y=Q in the 
original equation; and the ^-intercept by putting x = 0. 

62. Reduction of Ax + By + C = to the normal form. — 
In general A and B will not be the cosine and sine respectively 
of the same angle, and hence Ax + By + C = will not be in 
the normal form. In order to transform it to the normal form 
multiply both sides of the equation by an arbitrary constant 
k, whose value is to be computed later. 

This gives Akx + Bky + Ck = 0. 

The quantity k is now assumed to be such a number that 

Akx + Bky + Ck = 

will be identical with 

x cos 6 + y sin — p = 0. 

Comparing coefficients gives Ak = cos 6, Bk = sin 0, 
Ck = -p. 

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§62] EQUATION OF THE FIRST DEGREE 67 

To find the value of k, square both sides of the first two 
equations and add. This gives 

A 2 & 2 + B*k* = cos 2 + sin 2 - 1. 

Solving for A, k = j^—=- 

If & is replaced by its value, Akx + Bky + Ck = becomes 
Ax By C 

ran v 4. J 4. — — a 

1 J + VA 2 + B 2 T ±Va 2 + b 2 t ±VA 2 + b 2 

This is the general equation of the straight line expressed 
in the normal form. Either sign can be used with the radical, 
but of course the same sign must be used throughout the 
equation. Comparing this equation with the normal form 
gives 

[22] C0Se= ±VA^ + B^ Sine= ±VA° + B» 

P ±VA 2 + B 2# 

Hence, to transform the equation Ax + By + C = to 
the normal form, divide both sides of the equation by 

±Va* + b*. 

Example 1. — Change 3x — 4y + 6 = into the normal form. 

Here A - 3, B - -4, C - 6, and±VA? + £* - ±V9 + 16 - ±5. 

Dividing the equation through by ±5, it becomes 

±5 ±5 ^ ±5 " u * 

Either sign can be used since the two equations £ x — $ y + $ — 0, 
and — #£ + £y — £ =0are equivalent. 

EXERCISES 

Find the slope and y-intercept of the following equations by express- 
ing them in the slope intercept form. 

1. Zx + 2y - 4 - 0. 3. -5s + 2y - 6 = 0. 

2. 2x - 3y 4- 2 = 0. 4. 2s - 2y + 7 - 0. 



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68 



ANALYTIC GEOMETRY 



[563 



Change the following equations into normal form, and find the distance 
of the line from the origin. 

5. Zx - 4y - 6 - 0. 9. x + 2y - 3 - 0. 

6. -3s + 4y + 10 - 0. 10. -Zx + y - 6 - 0. 

7. 5x - 12y + 26 = 0. 11. 2x + 3 - 0. 

8. -5s - I2y + 39 = 0. 12. Zy - 4 - 0. 

Find the equations of the lines satisfying the following conditions: 

13. Through the point (1, 2) and parallel to Zx - 4y -f 6 - 0. 

14. Through the point (2, -3) and parallel to x + 2y - 3 - 0. 

15. Through the point (6, 2) and perpendicular to 2x + y — 3 = 0. 

16. Through the point ( —3, 1) and perpendicular tos — y + 6=0. 

63. Distance from a point to a line. — The aistance from 
the origin to the line x cos 6 + y sin 6 — p = 0, is the numeri- 
cal value of p. Hence if d! is the distance from the origin to 
the line Ax + By + C = 

where the sign of the radical is chosen so as to make d' positive. 



N 




Y' 



>i(*i.»> 



^ 



In order to find the distance d 
from the point Pi(xi, yO, Fig. 49, 
to the line. 4s + By + C = 0, 
translate the axes to the new origin 
P*(xi, yi). The equations of trans- 
lation [12] are 

x = x' + Xi, 
Fig. 49. » « »' + Vl* 

Making these substitutions, the equation 
Ax + By + C - 
becomes Ax' + By' + Axi + By x + C = 0, 
where the new constant term is 

Axi + By, + XJ. 
The distance d = PJt is the distance from the new origin to 
the line. From equation (1) 

m d . A* + g* + C 

1 J ±Va 2 



L 2 + B 2 ' 

where the sign of the radical is chosen to make d positive. 



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§63] EQUATION OF THE FIRST DEGREE 69 

Example 1. — Find the distance from the point (2, 1) to the line 

3s - 4y + 6 - 0. 

Solution. — Translate the axes so that the new origin is the point 
(2, 1). The equations of translation are 

s - s' + 2, 
y - v' + l. 
The equation of the line 3s — 4y + 6 =0, referred to the new origin is 

3(s' + 2) - W + 1) + 6 - 0, 
or 3s' - ±y' + 8 - 0. 

Putting this equation into the normal form gives 
3s' - V + 8 
±5 
Hence the distance from the new origin to the line is d ■* -£, 
and this is the distance from the point (2, 1) to the line 3s — 4y + 6 — 0. 
This distance could be found also by substituting directly in [28]. 
Putting A « 3, B « -4, C « 6, Xi - 2, y x - 1, 

, 3-2 -4-1 +6 +8 8 
±5 " +5 " 5* 

Example 2. — Find the distance from the point (3, —2) to the line 
5s + 12y - 4 « 0. 
Putting A - 5, B - 12, C - -4, xi - 3, and y & = -2 in [28], 
, 5-3 + 12(-2) -4 -13 

±13 " -13 * ' 

This apparent Inconsistency arses because both signs must first be 
put down and then the correct sign selected. 

EXERCISES 

Find the distances from the points to the lines in the following exercises: 

1. Point (2, 3) to line 4s - 3y + 4 - 0. 

2. Point (-1, 2) to line Sx + 4y - 6 - 0. 

3. Point (1, 3) to line x - y « 0. 

4. Point (2, 3) to line x cos 30° + y sin 30° - 3 = 0. 

5. Point (3, -1) to line x cos 135° + y sin 135° + 1-0. 

6. Point (1, 6) to line y - 1 - 3(s - 4). 

7. Find the altitudes of the triangle whose sides have the equations 
y = 1, 12s + 5y - 27 = 0, and 3s - 4y + 9 = 0. 

8. Find the altitudes of the triangle whose vertices have the coordi- 
nates (4, 2), (-3, 1), (6, -3). 



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70 



ANALYTIC GEOMETRY 



[§64 



64. The bisectors of an angle.— Let the sides of an angle be 
formed by the lines ST and SR, Fig. 50, the equations of which 

are A x x + B x y + C x = and 
{p(*v) A& + B%y+ Ct = respectively. 
Let P (x, y) be any point on the 
bisector SP of the angle formed 
by these two lines. 

From plane geometry it is 
known that the bisector of an 
angle is the locus of points equi- 
distant from the sides. 
Hence PT = PR. 




Fig. 50. 



Expressing this fact algebraically gives 

Axx + Bi y + d AjX + B % y + C 2 
±VA? 



5 + 5i 2 



±VA? 



+ B* 2 



The four possible combinations of signs in this equation will 
yield two different equations, 



[24] 



AjX + Bjy + Ci = A^x + Bay + C 2 

vV + Bx' - Va 2 * + b 2 * ' 



One of these is the equation of the bisector of the angle 
RST, while the other is the equation of the bisector of the 
supplementary angle TSR'. 

In order to tell which equation belongs to the bisector 
sought, draw the figure as accurately as possible and observe 
whether the slope of the required bisector is positive or 
negative. Since the two bisectors given by [24] are at right 
angles to each other, one has a positive slope and the other 
has a negative slope, so that in general it is easy to pick out 
the required equation. The exceptional case occurs when one 
bisector is very nearly parallel to the z-axis, and it is diffi- 
cult to tell the sign of its slope. In this case the numerical 
value of its slope is small, whereas the numerical value of the 
slope of the other bisector is large, so that again it is easy to 
associate the equations with the correct bisectors. 



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EQUATION OF THE FIRST DEGREE 



71 



Example. — Find the equation of the bisector of the angle which the 
line Zi = 3s + 4y — 5 = makes with the line 1% m bx — 12y + 6 - 0. 

In Fig. 51, let U be the required bisector. 

By [24] the equations of the two bisec- 
tors are 



3s -+ 4y - 5 



- ± 



bx - 112y + 6 




5 13 

Clearing of fractions and simplifying 
gives the two equations 

14* + ll2y - 95 - 0, (1) 

64x - Sy - 35 - 0. (2) 

The slope of (1) is small and negative, 

whereas the slope of (2) is large and 

positive. Since the slope of 1% is large 

and positive, its equation is 64x — Sy — 35 = 0. 

EXERCISES 

Find the equations of the bisectors of the angle which the first line 
makes with the second in exercises 1-6. 
1. 8* + y - 6 - 0, 7x + 4y - 3 - 0. 
x - 7y + 6 = 0, bx + by - 8 - 0. 
11* - 2y + 12 = 0, 2x + y - 6 = 0. 
13* + y - 15 - 0, 22* - Uy - 21 - 0. 
12* + Uy - 11 - 0, 9* - 2y + 10 - 0. 
9* + 7y - 6 - 0, 11* + Sy - 14 - 0. 

Find the equations of the bisectors of the angles of the triangle the 
equations of whose sides are 8* — y + 1 = 0, 
* + Sy + 1 - 0, and 7* 4- 4y - 43 - 0. 

8. Find the equations of the bisectors of 
the angles of the triangle whose vertices 
are flfc "¥), (1, «, and (12, -1). 

65. Systems of straight lines. — 

Sometimes the geometrical facts given 
are not sufficient to determine a 
straight line uniquely. In such a case 
not all the constants entering into 
the equation of the line will be determined. For instance, if 
the problem is to find the equation of a line that is parallel to 
3x + 4y : — 6 = 0, Fig. 52, it is evident that there are an un- 



2. 
3. 
4. 
6. 
6. 
7. 




Fig. 52. 



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72 ANALYTIC GEOMETRY [§66 

limited number of lines in the plane which satisfy the condi- 
tions of the problem. To find the equation of any one of these 
lines, substitute m = — f, in [16], which becomes 

y = — lx + 6, or 3x + 4y — 46 = 0. 

The quantity 6 can have any value whatsoever. If it is 
given some arbitrary value the equation 3x + 4y — 46 = 
becomes the equation of some one of the lines that are parallel 
to Sx + 4y — 6 = 0. All of these parallel lines taken together 
are said to form a system of lines. 

Another system of lines consists of all the lines through a 
given point. If the point has the coordinates (1, 2), the equa- 
tion of this system of lines is y — 1 = m(x — 2), by [15]. 

EXERCISES 

Find the equations of the following systems of lines: 

1. All the lines passing through the point (—2, 3). 

2. All the lines passing through the origin. 

3. All the lines passing through the point (3, 4). 

4. All the lines having their x-intercept equal to 3. 
6. All the lines haying their y-intercept equal to —4. 

6. All the lines at a distance 3 from the origin. 

7. All the lines at a distance 7 from the origin. 

8. All the lines parallel to the line 2x + y — 3 = 0. 

9. All the lines perpendicular to the line x — Sy +' 6 =» 0. 

10. All the lines such that the a?-intercept of each is equal to its 
y-intercept. 

66. Applications of systems of straight lines to prob- 
lems. — Sometimes the facts determining a straight line are 
not such that its equation can be written down immediately. 
This happens if the slope m and the distance p of the line 
from the origin are given. In such a case there are two 
methods of procedure, one is to compute the constants which oc- 
cur in some standard form of the equation of a straight line, by 
drawing the figure and applying plane geometry or trigonome- 
try. Another method is illustrated in the following example. 



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§66] 



EQUATION OF THE FIRST DEGREE 



73 



Example. — Find the equation of the straight line given m - J, and 
p = 3. 

First method. — First write down the equation of the system of lines 
whose slope is t . This is y « |x + b. 

Next transform this line into the normal form. Its equation becomes 
4x - 3y + 3& 
±5 
The distance of this line from the origin is 



3. 



3, and b 




36 
77, but this is p and p 

Hence -r-= * 

±5 

Substitute this value of 6 in y 
and it becomes y = Jx ± 5, 
or 4x - 3y ± 15 - 0. 

There are two lines which satisfy the 
required conditions, and they are equally Fig. 53. 

distant from the origin. 

Second method. — Write down the equation of all lines distant 3 from 
the origin. This equation is x cos $ + y sin $ — 3 — 0. In order to 
determine 0, note that the slope of this line is — cot 0. 

Therefore — cot $ = $ , and $ can be in either the second or the fourth 
quadrants. 

If $ is in the second quadrant, then sin $ « f and cos $ « — 1« 

If 6 is in the fourth quadrant, then sin $ — — i 
and cos 6 =» |« 

Substituting these values, the equation be- 
comes ± \x T $ y — 3 - 0. 

Multiplying both sides by ±5, gives 
4x - Zy ± 15 - 0. 

Example 2. — Find the equation of a line 
^*-£ through the point (1, 3) and making equal 
intercepts on the axes. 
First solution, geometric method. — In Fig. 54, 
Fig. 54. let AB be the line through Pi(l, 3) whose equa- 

tion is to be found. Since the intercepts are 
equal, angle BAO » angle OB A = 45°. 

Hence a = OM + MA - OM + MP X -1+3-4, and this is 
also the value of 6. 

x 1/ 
Therefore the required equation ^4 + 4 = 1, orx + y — 4. 

Unfortunately by using the geometric method parts of the solution are 




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74 ANALYTIC GEOMETRY [§66 

liable to be overlooked. This is illustrated very well in this problem, 
since the line OP\ passing through the origin and the point (1, 3), satisfies 
all the conditions of the problem and is therefore also a solution. 

Second solution, algebraic method. — Since the line AB passes through 
the point (1, 3) its equation is y — 3 ■= m(x — 1). The ^intercept 

of this line is — — — > and the ^-intercept is 3 — m. 

Since these are equal, — - — » 3 — m. 

Solving this equation gives m = 3 or — 1. 

If to — 3 the equation is y — 3 = Z(x — 1) or y — Sx = 0. 

If to = — 1 the equation is y — 3 = — (x — 1) or £ + y — 4=0. 

ThircTsotution, algebraic method. — This differs from the preceding only 
in that it starts from the intercept form of the equation of a straight 
line, instead of from the point slope form. 

Since the intercepts are equal, the intercept form of the equation is 

- + - = 1. In order to make this line pass through the point (1, 3), 

1 3 

substitute these coordinates for x and y. This gives — h - = 1. 

a ■ a 

Solving,, gives a — 4 and the required equation is j + | — 1, 

or x + y = 4. 

The question naturally arises, what happened to the solu- 
tion y — 3x = 0? This is certainly a solution since the 
intercepts a = and b = are equal and the line passes 
through the point (1,3). This question can be answered by 
noting as stated in article 57 that the intercept form is not 
valid when either or both intercepts are 0. Hence the solution 
y — 3x = cannot be secured from the intercept form. 
Whenever this form of the equation of a straight line is used 
the question as to whether either or both intercepts are zero 
must be answered independent of the equation. 

EXERCISES 

Find the equations of the lines determined by the following conditions: 

1. The slope of the line equals — } and it is distant 1£ units from the 
origin. 

2. The line makes equal intercepts on the axes and passes through the 
point (4, 2). 



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§67] 



EQUATION OF THE FIRST DEGREE 



75 



3. The line passes through the point ( — 7, 4) and is tangent to a circle 
whose center is the origin and radius equal to 1. 

4. The line passes through the point (4, 2) and is tangent to a circle 
whose center is the origin and radius equal to 2. 

6. The slope of the line is 2 and its z-intercept equals 3. 

6. The slope of the line is —2 and the sum of its intercepts is 9. 

7. The slope of the line is — } and the sum of its intercepts is 5.. 

3. The line makes intercepts which are equal numerically but opposite 
in sign, and passes through the point (6, 3). 

9. The line passes through the point (1, 3) and the sum of its intercepts 
equals 8. 

10. The line passes through the point (3, 1) and the portion in- 
cluded between the axes is bisected by this point. 

11. The line passes through the point (3, y/Z) and the perpendicular 
from the origin on the line has an inclination of 60°. 

12. The line is perpendicular to the line 4z + 3y — 6 — and distant 
2 units from the origin. 

13. The line is distant 3 units from the origin and its y-intercept 
equals 5. 

14. The line is distant 2 units from the origin and the product of its' 
intercepts is y. 

16. The line passes through the point (1, 2) and makes with the axes 
a triangle in the first quadrant whose area equals 4. 

16. The line passes through the point (1, 2) and makes with the axes 
a triangle in the second or fourth quadrants whose area equals 4. 

67. Loci through the intersection of two loci. — Theorem. 
Iff(*> V) = and g(x, y) = are 
the equations of any two loci and Y 

k is any constant not zero, then o(*.v)^o^ 
f(*> y) + tyix, y) = is the equa- 
tion of a curve which passes 
through all the points of inter- 
section of f(x, y) = and 
9(&> y) — 0, but does not intersect 
these curves in any other point 

Proof. — Let Pi(xi, j/i), Fig. 55, be any point of intersection 
of f(x, y) = and g(x, y) = 0. Since Pi lies on both these 
curves its coordinates must satisfy each equation, therefore 
f(x u yd = and g(x u y x ) =0. 




/r*.v)-o 



*-* 



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76 ANALYTIC GEOMETRY [§67 

Substituting the coordinates of Pi in /(a, y) + kg(x, y) = 0, 
/(*it Vi) + kg(x lt y x ) - + fc = 0. 

Therefore Pi lies also on the curve f(x, y) + kg(x, y) = 0. 

But Pi was any point of intersection of f(x, y) = and 
Q&i y) a 0, therefore every point of intersection of these 
curves lies on f(x, y) + kg(x, y) = 0. 

Furthermore the curve /(x, y) + kg(x, y) = cannot meet 
either f(x, y) = or g(x, y) = in any other point. For if it 
did, suppose it meets /(x, y) = at Pj(x2, 2/2), and that P2 is not 
on g(x, y) = 0. Then /(x 2 , j/2) = 0, but g(x 2 , #2) = a where 

Substituting the coordinates of P 2 in /(a;, y) + A#(a;, y) = 0, 
/(*a, y«) + kg(x t , y 2 ) = + ka 5* 0. 

In like manner it can be shown that f(x y y) + fcgrfo y) = 
will meet g(x, y) = only at the points of intersection of 
/(a, y) = and y(x, y) = 0. 

If /(x, y) = is the straight line Ax + By + C = 0, and 
g( x > y) — is the straight line it's + B'y + C" = 0, then 

/Or, y) + fcg(x, y) = 0, 
or Ax + By + C + k(A'x + B'y + C") = 

is the equation of a straight line through the point of in- 
tersection of the straight lines, Ax + By + C = and 
A'x + B'y + C" - 0. 

Example 1. — Find the equation of the straight line which passes 
through the point (4, 3) and through the intersection of the two lines 
2x + 3y - 5 - and 3s - 4y + 1 - 0. 

It has just been shown that the equation of any line passing through 
the intersection of these two lines is of the form 

2x + Zy - 5 + k(Zx - 4y + 1) - 0. 
Since this line passes through the point (4, 3), its equation is satisfied 
when x — 4 and y = 3. This gives 

12 + fc(l) - 0. 
Therefore A; «■ —12, and the required equation is 

2x + 3y - 5 - 12(3* - 4y + 1) - 0, 
or 2s - Sy + 1=0. 



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§68] 



EQUATION OF THE FIRST DEGREE 



77 



EXERCISES 

Find the equations of the lines satisfying the following conditions. 

1. Passing through the point of intersection of 2x + Zy — 3 — 
and Zx — y — 1 — 0, and through the point (1, 1). 

2. Passing through the point of intersection of 5x — 4y — 2 — 
and 2x + 4y — 15 ■» 0, and through the point (2, 3). 

3. Passing through the point of intersection of Zx + 2y — 6 — 
and 3 + y ™ 3, and perpendicular to 2a; + !/ — 1 ■*(). 

4. Passing through the point of intersection of x — 6y — 3 and 
2a: — y *■ 2, and perpendicular to 3 — 2y + 1 «■ 0. 

6. Passing through the intersection of y - 6 + x and 3y — 4 — 2x, 
and parallel to x + Zy — 4 — 0. 

68. Plotting by factoring. — Since it is easy to plot a straight 
line, the theorem of article 48 gives a simple method of plotting 
equations which can be factored into linear factors. 
. Example. — Plot the equation 2x* + 2x + 7y - xy + Zy 1 + 4. 

. First transpose all terms to the left hand side of the equation 
2x* + 2x + 7y - xy - 3y* - 4 - 0. 

In order to find out if this equation 
can be factored, regard it as a quadratic 
in x or y, and solve for that variable. 
For the sake of convenience the variable 
chosen this time will be x. Collecting 
like powers of x, 

2s* + (2 - y)x - 4 + 7y - 3y* - 0. 

Solving for x, by means of the formula, 
Art. 4, where a = 2, 6 ■» 2 — y, and 
<j a -4 + 7y - 3y*, 

± V25y« - 60y + 36 _ -2+y ± (5y - 6) , 




*-x 



Fiq. 56. 



'- 2 + y 



4 4 

Hence x — -^ — or — y + 1, and the left hand side can be factored 

in to2(x-^li)( a;+y -l). 

The equation now becomes (2x — Zy + 4)(x + y — 1) « 0. 
Therefore the graph of 2s* + 2x + 7y = ^ry + 3y* + 4 consists of 
the two straight lines 2x — Zy + 4 = and x + y — 1 - 0. 

When an equation of the second degree in each of two 
variables, is solved for one variable in terms of the other, an 



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78 ANALYTIC GEOMETRY [§69 

expression is obtained under a radical sign. If this expression 
is a perfect square, the graph of the equation consists of two 
straight lines. 

Thus, in the problem just solved, 25y* — 60y + 36 is a perfect square. 
If it had not been a perfect square 2x* + 2x + 7y — xy + 3y* + 4 could 
not have been plotted by this method. 

Example 2. — Plot the curve x*y =* y». 

Transposing all terms to the left hand side, x*y — y» = 0. 
• Factoring the left hand side, y{x — y)(x + y) — 0. 

The graph consists of the line y = which is the x-axis, the line 
x — y = o and the line x + y » 0. 

EXERCISES 

1. Find the equation of the triangle whose sides are x — y f y — 0, 
and x + y = 1. 

2. Find the equation of the square whose bounding lines are x *= 1, 
i»2,y»l, and y = 2. 

Plot the following curves by first factoring: 

3. rr* - 2y* - *y + Sy - 1 - 0. 

4. 2y* = xy + x*. 

6. x 2 + 2x + 1 - 4y*. 

6. 2x* + xy + 4x + y + 2 - y*. 

69. Straight line in polar coordinates. — In general the 
equations of straight lines in polar coordinates are not as 

simple as those in rectangular co- 
ordinates. The simplest case is the 
one in which the known quantities 
are the polar coordinates of the 
foot of the perpendicular from the 
origin to the line. This is the same 
data as was given for the normal 
form of the equation of a straight line. If the end of the 
perpendicular to the line from the origin has the coordinates 
(pi, 0i), Fig. 57, let P(p, 6) be any point on the line. 
Then ' gp 

cos POPi = nff> or cos (e - *i) = ~ 




Hence 



OP ' ~ l WD vv V1/ P 

- 90 = px. (1) 



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§70] EQUATION OF THE FIRST DEGREE 79 

This equation can also be obtained from the normal form 
of the equation of a straight line by replacing p by p l9 by 0i, 
x by p cos 0, and y by p sin 0. 

In the special case where the line is perpendicular to the 
polar axis, B x = 0, and the polar form of the equation of the 
straight line takes the form p cos = pi. If the straight 
line is parallel to the polar axis, 0i = 90° and the equation of 
the straight line becomes p sin = p x . 

Example. — Find the polar form of the equation of a straight line if the 
coordinates of the foot of the perpendicular drawn to it from the origin 
are (3, 60°). 

Substituting in equation (1), gives p cos (0 — 60°) = 3. 

EXERCISES 

Write the equations of the following straight lines in polar coordinates, 
if the coordinates of the end of the perpendicular from the origin to the 
line are: 

1. (3, 45°). 3. (7, 90°). 6. (-4, 135°). 

2. (-2, 60°). 4. (4, 180°). 6. (3, 315°). 
Change from rectangular to polar coordinates. 

7. x + y - 1 - . 1 . xVS + y - 4. 

8. x - 3. 11. x - yy/3 +6=0. 

9. y = -7. 12. y - 2x - 0. 
Change from polar to rectangular coordinates. 

18. p = 3 sec 0. 18. p - -. — w—. — - • 

4co8 0-6 8in0 

14. p = 4 esc 6. 19. 5 sin 6 — 3. 

15. tan $ - 6. 20. 13 cos 6 = -5. 
1ft 2 V2 



cos + sin p sin (6 - 45°) 

4IT 3(cos 6 — sin 0) fto 3 

17 » P * — o/i **• 



cos 20 K cos (0 + 60°) 

70. Applications of the straight line. — Whenever two vari- 
ables are related so that one varies directly as the other, or 
so that a change in one varies directly as the corresponding 
change in the other, the relation between the variables is 
linear, and the graph showing the relation between the 
variables is a straight line. 

Since many of the relations in physics, mechanics, and 



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80 ANALYTIC GEOMETRY [§70 

engineering are of this nature, the straight line has a wide 
field of application. Oftentimes the curves representing the 
relation between physical quantities are within certain limits 
so nearly straight lines that the more complicated equation 
is replaced, on account of its simplicity, by the linear relation. 
A few specific instances are the following. 

(1) The increase in velocity of a body falling under the 
action of gravity is proportional to the time. This is expressed 
by the relation 

v — t> = k(t — Jo), 
where vo is the velocity of the body at the time t and v is the 
velocity of the body at any time t. If v and v are expressed 
in feet per second, and t and to are expressed in seconds, 
then k, the proportionality factor, is the familiar constant g. 
This relation is often expressed 

v = kt + v 0f 
where v is the velocity when t = U = 0. 

(2) Hooke's Law. — The extension of an elastic string varies 
directly as the tension. This is expressed by the relation 

I = kt + Jo, 
where I is the length of the string under the tension t y and U is 
the length of the string when t = 0. 

(3) The expansion of a bar due to heat, is very nearly pro- 
portional to its increase in temperature. This is expressed 
by the relation 

I — Z = k(t — t ), 
where h is the length of the bar at some temperature t Q and 
I is its 1 length at any temperature t 

(4) The weight of a column of mercury in a barometer 
varies directly as its height. This is expressed by the relation 

10 = kh, 
where the weight w is taken as zero when h = 0. 

In all four cases the graph representing the linear relation 
between the variables is a straight line. 



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§70] EQUATION OF THE FIRST DEGREE 81 

For further applications of the straight line see Chapter X 
on empirical equations. 

GENERAL EXERCISES 

1. Translate the following algebraic statements into words and draw 
their loci: 

(1) V - 4s, (2) y - 4* - 4, 

(3) x - 3y + 2, (4) x - 5y - 2. 

2. Find the equation of the line (1) through the point (4, —3) and 
parallel to 2x — 3y — 4; (2) through the point (5, 7) and perpendicular to 
2x + 7y - 14. 

3. Find the equation of the line (1) through the point (—2, —5) 
and parallel to x — 7y — 3; (2) through the point (h, k) and parallel 
to the line y = mx + 6. 

4. Find the length of the following perpendiculars: 

(1) From (3, 2) to 4x - 3y - 7 - 0. 

(2) From (0, -3) to 5x - y - 6 - 0. 

(3) From (2, 3) to Ox - % - 10 - 0. 

6. Find the lengths of the three altitudes of the triangle whose- 
vertices are (4, 5), (—2, 2), and (3, —4). 

6. Find the distances from the line 2x + 3y — 12 = to each of the 
points (4, 4), (2, -3), (0, 0), (-3, 5), and (-2, 8). 

7. Given 4x -f ky — 5 =0; determine the value of k for which the 
line will (1) pass through the point (—4, 3), and (2) be parallel to 
3* - 2y + 7 - 0. 

8. Find the equations of the lines through the intersection of the lines 
2s -|- y — 16 = and x — y -f 2 = and also 

(1) passing through the point (2, 7), 

(2) parallel to the line 7x - 2y + 6 = 0, 

(3) perpendicular to the line 3a; — 4y + 2 = 0, 

(4) having the slope ~f. 

9. Given a triangle having as vertices the points (6, 2), (—3, 5), 
and ( — 1, —3); find the equations of the perpendicular bisectors of the 
three sides, and the coordinates of their point of intersection. 

10. Show that 15a:* — 14xy — 8y* = is the equation of two straight 
lines intersecting at the origin. 

11. Prove that, if A, B, and C are real numbers, Ax* + Bxy -f Cy* - 
represents two straight lines passing through the origin, and that these 
lines are real and distinct, real and coincident, or imaginary according 
as B* — 44 C is positive, zero, or negative. 

6 



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82 ANALYTIC GEOMETRY [§70 

12. The perpendicular drawn from the origin to a line makes an angle 
of 60° with the s-axis and its length is 2, find the equation of the line. 

13. Write the equations of the following lines: 

(1) Passing through the point (3, 5) and having an inclination of 45°. 

(2) Passing through ( — 1, —3) and having a slope of 2. 

(3) Passing through (—2, 8) and having an inclination of 120°. 

14. Show that the following lines form a parallelogram: 

(a) 2s + 3y - 10, (6) 2s + Zy - 20, (c) s - 2y - 5, 

(d) 2x - 4y - 17. 

15. Write the equation of the line passing through the intersection 
of x — 3y + 8 = and 3s -f 2y + 2 = and making an angle whose 
tangent is 2 with the s-axis. 

16. Find the coordinates of the point in which the perpendicular to 
the line 2x — y — 1 = and passing through ( — 2, 3) intersects that 
line. 

17. What does the equation 3s — 2y + 4 =0 become when the 
coordinate axes are turned through an angle of 45°? Plot the locus of 
the equation in both cases. 

18. Plot each of the following lines, translate the axes so that the new 
origin shall be at the point indicated and replot from the new equation. 

(1) y = 3s + 4, (2, 3). (3) y - mx + 6, (c, d). 

(2) 2y - 3s - 2 - 0, (-2, 3). (4) y - 4s + 5 - 0, (}, -2). 

19. The three vertices of a triangle are (8, 2), (4, 8), and (— 2,-6). 
Find the equations of the lines each of which bisects two sides of the 
triangle. 

20. Given two straight lines each having an inclination of 45° and 
having intercepts on the y-axis of 6 and —8 respectively; find the equa- 
tion of the straight line that is equidistant from the two lines. 

21. Find the equation of a straight line such that the perpendicular 
from the origin to it equals 8 and makes an angle of 45° with the s-axis. 

22. Find the equation of the straight line which passes through the 
intersection of the lines s — 2y — 4 = and s + 3y — 8 = and is 
parallel to the line 3s + 4y = 4. 

23. Find the equation of the line through the point (2, 3) making an 
angle tan" 1 } with the line 2s — 4y + 7 = 0. 

24. Find the equation of the line through the point ( — 1, 2) making 
an angle sin" 1 J with the line s + 3y — 4 — 0. 

25. Find the equation of the line through the point (6, 4) making an 
angle cos" 1 (—J) with the line 2s — y + 6 — 0. 

26. Find the equation! of the two lines through the point ( — 1, —3), 
which form an equilateral triangle with the line s + y » 2. 



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§70] EQUATION OF THE FIRST DEGREE 83 

27. Find the equation of the line through the point (0, 6) which together 
with the {/-axis as the other equal side forms an isosceles triangle with 
the line 2x — y + 4 = 0. 

28. Find the equation of the line through the point (0, 6) which together 
with the 2/-axis for the other leg forms an isosceles triangle with the line 
2x + y - 4 = 0. 

29. The equations of the two equal sides of an isosceles triangle are 
x — 2y + 6 — and 2x — y — 2 = 0. Find the equation of the third 
side if it passes through the point (9, 4). 

30. Find the slope of the line 2x + 3y — 4 = after the axes are 
rotated through 30°. 

31. Find the slope of the line x — Zy + 6 = after the axes are rotated 
through the angle 0, where cos 6 — — J and 0is in the second quadrant. 

32. Find the equations of the two lines through the point ( — 1, 3) 
which trisect that part of the line 2x + y — 6 — which is intercepted 
between the axes. 

33. An equilateral triangle lies wholly in the first quadrant. If one 
side has its extremities at (1, 6) and (6, 1), what are the equations of the 
other two sides? 

34. An isosceles right triangle is constructed with its hypotenuse along 
the line 2x + y — 6 = 0. If its vertex is the point (3, 4), find the 
equations of its sides. 

36. A circle is inscribed in the triangle the equations of whose sides 
are x + 2y - 16 = 0, 2x - y + 3 = 0, and 2x + y - 7 - 0. Find 
its radius and the coordinates of its center. 

36. The base of an isosceles triangle is the line joining the points (1, 5) 
and (4, 6), its vertex is on the line x + y — 7 = 0. Find the codrdi- 
nates of its vertex. 

37. Find the locus of a point which moves so as to be always equi- 
distant from the points (3, 5) and ( — 1, 7). 

38. Find the equation of the locus of a point which moves so that 
its distance from the line 7x + 4y — 6 — is twice its distance from 
the line x - Sy + 3 = 0. 

39. Find the equation of the locus of a point which moves so that 
the difference of the squares of its distances from the points (—2, 3) 
and (1, 6) shall be constant and equal to 2. 

40. Find the equations of two lines through the point (1,1) such that 
the perpendiculars let fall from the .point (1, 3) on them are each of 
length f . 

41. Prove that the feet of the perpendiculars let fall from the point 
(3, 1) on the sides of the triangle x = 0, y =» 0, and 2x + y — 4 = 
lie in a straight line. 



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84 ANALYTIC GEOMETRY [§70 

42. Find the equations of the straight lines through the point (3, 6) 
and intersecting the line x+y — 2=Q&t& distance 5 from this point. 

48. Prove that the perpendicular bisectors of the sides of a triangle 
meet in a point. 

44. Find the equation of the locus of a point that is always twice as 
far from the origin as from the s-axis. 

46. The coordinates of two points are (3, 5) and (4, 4). Find the 
equation of a straight line which bisects the line segment connecting these 
points and makes an angle of 45° with the s-axis. 

46. A straight line inclined to the s-axis at an angle of 150° has an 
s-intercept equal to 8. Find the equation of a straight line passing 
through the origin and bisecting that portion of the line included between 
the axes. 

47. Find the equations of the four sides of a square two of whose 
opposite vertices are (2, 3) and (3, 4). 

48. A straight line moves so as to keep the sum of the reciprocals of 
its intercepts on the axes a constant. Show that the moving line passes 
through a fixed point. 

49. Find the equation of the straight line passing through the point 
(2, 6) and making an angle of 30° with the line s — 2y » 1. 

60. Find the equation of a straight line passing through the point 
(c, 0) and making an angle of 45° with the line bx — ay = ab. 

61. The equation of a straight line is 3s + by = 15; find the equa- * 
tion of the same line referred to parallel axes whose origin is at (3, 2). 

62. Find the equations of the straight lines bisecting the angles formed 
by the lines 12s + 5y = 8 and 3s — 4# = 3. 

68. Show that an angle of 45° is formed by the lines represented by 
the equation x* — xy — %* + 2x — y + 1 =0. 

64. Given the equation Ax + By + C — 0. Find the relation be- 
tween A, B and C, (1) so that the s- and y-intercepts shall be equal; 
(2) so that the inclination of the line shall be 45°; (3) so that the line 
shall pass through the point (1, 2). 

66. Determine the angle that the first line of each of the following 
pairs makes with the second: 

(1) x + 2y = 5, 3s - 4|/ - 4. 

(2) 3s + 4y - 6, 2x - y - 2. 

(3) VZx + y - 4, V3s - y + 4 = 0. 

66. Determine the value of m in y = ma; + 6, so that it shall make 
an angle of 60° with x - 2y = 3. 

67. Find the coordinates of the point through which the three lines 
y — 4x - 5, y ] — 3s - 4, and y — 2s = 3 pass. 



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§70] EQUATION OF THE FIRST DEGREE 85 

68. Find the value of m so that y — mx + 3 shall pass through the 
intersection of y — z = 1 and y — 2x = 2. 

69. Find the equation of the line perpendicular to 5x -f 8y — 3 and 
having a y-intercept equal to 6. 

60. Find the angle which the line 4a; — y - 8 makes with the line 
&r - y - 9. 

61. Find the equation of the locus of a point whose distance from 
3x -f 4y s 5 is one-half its distance from 12s — 5y — 16. 

62. Given the two fixed points Pi(-2, 4) and Pj(l, 3). Find the 
equation of the locus of the variable point P(x, y) which moves so that 
the area of the triangle PP\P% is always equal to 10. 

63. Find the equation of the locus, of a point which moves so that 
the slope of the line joining it to the point (0, 2) is twice the slope of the 
line joining it to the point (0, —2). 

64. If the equations of the sides of a triangle are x + 2y — 15-0, 
2x — y + 5 = 0, and 2x — lly + 15 = 0, find the coordinates of the 
point of intersection of the bisectors of the interior angles of the 
triangle. 

66. Find the equation of a line passing at a distance y/2 from the 
origin if the sum of its intercepts is 4. 
66. If the three lines 

A x x + B x y + Ci - 0, 
A t x + B 2 y + C, + 0, 
A,x + B*y + C t = 0, 
meet in a point, show that 

Ai Bi C x 

A t B % C 2 = 0. 

A z Bi C t 



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CHAPTER V 

THE CIRCLE AND CERTAIN FORMS OF THE SECOND 
DEGREE EQUATION 

71. Introduction. — The circle affords other examples of the 
ease and power obtained in analytic geometry by applying 
algebra to geometry. Since the properties of the circle are 
well known from plane geometry, atten- 
tion can be confined to the methods 
>P(*jt) used in solving the various problems. 

72. Equation of circle in terms of 

center and radius. — A circle is defined 

+ x in plane geometry to be the locus of all 

points in a plane equidistant from a 

fixed point in the plane called the center 

Fio. 68. °f the circle. 

Let the center of the circle be the fixed 
point, C(h, k), Fig. 58, and let the constant distance, or 
radius, be r. 

Then if P(x, y) is any point on the circle, the distance 
PC - r. 

But by [3], PC - V(g - hy + (y - k) \ 
Then \/(x - h)* '+ (y - k) 2 = r. 

[25] .\(x-h)* + (y-k) 2 = r 2 . 

Furthermore, comparison of this equation with [3] shows 
that every equation of the form of [25] is the equation of a 
circle. 

If the center of the circle is the origin, this equation takes 
the simple form 
[26] x 2 + y* = r 2 . 




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§73] THE CIRCLE AND CERTAIN FORMS 87 

73. General equation of the circle. — Equation [25] when 
expanded becomes 

x 2 + y 2 - 2hx - 2ky + h 2 + k 2 - r 2 - 0. 

This is in the form 
[27] x 2 + y* + Dx + Ey + F = 0. \S 

This is called the general equation of the circle. 

Conversely f every equation in the form of [27] is the equa- 
tion of a circle, since after completing the squares in the x and 
the y-terms, it can be written in the form 

2)2 £2 £)2 £J2 

x 2 + Dx + — + y 2 + Ey+ _ . _ + _ _ F| 

or (* + \D) 2 + (y + PO 2 = {WD 2 + E 2 - 4F) 2 . 

Comparison of this equation with e quation [25], sh ows that 
h = -iD, k = -\E, and r = WD 2 + E 2 - 4F. 

Therefore every equation in the form of [27] is the equation 
of a circle. 

If D 2 + E 2 — 4F>0, equation [27] represents the equation 
of a real circle. 

If D 2 + E 2 - 4F = 0, the radius of the circle equals 0, and 
the locus becomes a point. Such a circle is called a null or 
point circle. 

If D 2 + E 2 — 4jF<0, the radius of the circle is imaginary 
and the circle is called an imaginary circle. 

74. Special form of the general equation of the second 
degree. — The equation of a circle is a special case of the most 
general equation of the second degree in two variables 

Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. 
In order that this shall be the equation of a circle comparison 
with [27] shows that B = and A = C, for then this equa- 
tion becomes 

Ax 2 + Ay 2 + Dx + Ey+F = f 
which can be reduced to [27] by dividing by A. The quantity 
A cannot be zero, since if it were, this equation would become 
the equation of a straight line. 



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88 ANALYTIC GEOMETRY [575 

Example. — Find the codrdinates. of the center, and the radius of the 
circle 6x» + 5y» + 2x - 3y - 4 - 0. 
Solution. — Dividing by 5. 

** + V* + ix - |y - f « 0. 
Completing squares 

*■ + 1* + * + »■ - f v + tH - A + tH + f 

or (x + *)* + (y - A) 1 - tWt- 

Comparing this equation with [26], shows that the center has the 
codrdinates (—J, tV) an( ^ that the radius is equal to VSt = t^tV^S- 

This problem could also be solved by substituting the values D — $ , 
J0 — -f , and F » -f , in the formulas of Art. 73. 

EXERCISES 

Find the codrdinates of the centers and the radii of the following circles: 

1. x* + y* - 2x - 4y - 4 - 0. 

2. x* + v* + 4s - 6y + 12 - 0. 

3. x* + y* + 12* ' + by + 41 - 0. 

4. ** + y* - x - 4y + 2 - 0. 

5. 3x* + 3y* - 2* - 4y + 1 - 0. 

6. 2x* + 2y* + x + 3y - 5 - 0. 

7. 2s* + 2y* + 2x + 6y + 5 - 0. 

8. a; 1 + y* - 2os - bay + a* « 0. 

9. 2x* + 2y* + 12ax + lOay - a* - 0. 
10. 9x* + 9y* - 6aa? + 15ay + 6a* - 0. 

75. Equation of a circle satisfying three conditions. — Since 
both equations [26] and [27] involve three arbitrary constants, 
the circle is determined if enough geometric or algebraic con- 
ditions are given to determine the three constants uniquely. 

There are two methods of procedure. One is to compute 
the constants in [25] geometrically. That is to say, from the 
given conditions compute the radius of the circle and the 
codrdinates of its center, then substitute these values in [25]. 
Another method is to set up three equations involving h, k 9 
and r, or three equations involving D, E, and F, and solve 
these equations simultaneously. This method is generally 
more satisfactory, and is illustrated for both sets of con- 
stants in the following example. 



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575] THE CIRCLE AND CERTAIN FORMS 89 

Example 1. — Find the equation of a circle passing through the points 
(3, 5), (4, 4), (1, 1). 

First method, geometrical. — Find the equations of the perpendicular 
bisectors of two of the sides of the triangle (3, 5), (4, 4), (1, 1). 
Solve these equations simultaneously. This gives the coordinates of 
the center of the circle. 

Next find the distance from the center of the circle to any one of the 
three vertices. This gives the radius of the circle. Substituting 
the values of h, k and r thus found in [26] gives the desired equation. 
It is obvious that this method is long and hence the actual computa- 
tion is not given. A shorter method is the following. 

Second method, algebraic. — Make the coordinates of each of the three 
points satisfy the equation (x — h) % + (y — *)* ■ **• Tk* 8 8* ve8 
(3 _ h) i + ( 5 _ *)» - r«, * 
(4 - *)* + (4 - *)« - r«, 
(1 - A)« + (1 - *)* - r*. 
Simplifying each of these equations gives 

h* + &* - 6fc - 10* - r« + 34 - 0, 

h* + h* - 8A - 8fc - r* + 82 - 0, ' 

A* + *« - 2h - 2* - r* + 2-0. 

Solve these equations by subtracting the second from the first and 

the third from the second. Then solving the two equations thus 

obtained gives h — 2, k = 3, r — y/b. 

Hence the equation of the required circle is 

(x - 2)* + (y - 3)* - 5 
Simplifying, this becomes 

x* + y* - 4s - 6y + 8 - 0. 
Third method. — Make the coordinates of the three points satisfy 
the equation x % + y* + Dx + Ey + F « 0. This gives 
9 + 25 + 3D + SE + F - 0, 
16 + 16 + 4Z> + 4E + F - 0, 

Solving these equations simultaneously gives 

D - -4, JE - -6, F - 8. 

Hence the equation of the circle is 

x s + y»- 4x - 6y + 8 - 0. 

Example 2. — Find the equation of a circle which passes through the 
points ( — 1, 7) and (7, 1) and is tangent to the line x + y — 10 - 0. 



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90 



ANALYTIC GEOMETRY 



[§75 



This problem illustrates how a combination of both algebraic and 
geometric methods may sometimes be useful. 

Solution. — Use the equation (x — A) 1 + (y — k)* — r f , and make the 
circle go through the points ( — 1, 7) and (7, 1). This gives the two 
equations (-1 - A)« + (7 - k)* » r«, (1) 

and (7 - A)« + (1 - ib)« =r«. (2) 

Since the line x + y — 10 = is tangent to the circle, the distance 
from the point (A, k) to the line x + y — 10 « equals r, hence by [23] 

Y A + k - 10 



\ 


I 






v^foS 












(Wk 


X w tr 




o 




\5 











(3) 
like 



±V2 
Simplifying and combining 
terms in (1) and (2) gives 

A* + ib* + 2A - 14& + 50 « r*, (4) 
A* + fc* - 14A - 2fc + 60 - r*. (5) 

Subtracting (5) from (4) and divid- 
ing both sides of the resulting equa- 
tion by 4, 

4A - Zk - 0. (6) 

Substituting the value of r from 
equation (3) in (4) and simplifying, 

A* - 2hk + k* + 24A - 8k - 0. (7) 
Substituting A = £ k from equation (6) in equation (7) and simplifying, 

k* + 160fc = 0. 
Hence k = or h = -160. 

Computing the value of A from equation (6), gives A = or A — — 120. 
Computing the value of r from equation (3), gives 

r = 25\/2 orr = 145\/2. 
Substituting the values of A, &, and r in the general equation of the 
circle gives the two solutions x* + y* — 60, 



Fig. 59. 



and 



(x + 120)* + (y + 160)* - 42,050. 



EXERCISES 

Find the equations of the circles through the following points: 



1. (5, 5), (1, 3), (2, 6). 

2. (3, -2), (-1, -4), (2, 

3. (0,3), (-4,3), (-3,4). 

4. (1,6), (4,5), (-3, -2). 

5. (-1,4), (-4, -5), (3,2). 



6. (2,8), (-1, -1), (-2,6). 

5). 7. (2, 4), (2, -2), (3, 3). 

8. (4,4), (5,3), (-3,3). 

9. (3, 7), (1, 1), (5, 3). 
10. (-1,1), (1,5), (-5,3). 



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§75] THE CIRCLE AND CERTAIN FORMS 91 

Find the equations of the circles fulfilling the following conditions:' 

11. Passing through the origin, radius 5, and ordinate of center —3. 

12. Passing through the origin, radius 13, and abscissa of center 12. 

13. Center at origin and tangent to line x + 2y — 10. 

14. Center at point (1, 2) and passing through the point (3, — 1). 
16. Center at ( — 1, 3) and tangent to line Zx + y — 10 — 0. 

16. Center on x-axis and passing through the points (3, 3) and (5, —1). 

17. Radius 5 and passing through the points (5, 6) and (2, 7). 

18. Radius 5 and tangent to the line 4x + Zy «- 16 = at the point 
(1, 4). 

19. Having the line joining (—3, 2) and (5, 6) as diameter. 

20. Passing through the point (1, 1) and having the same center as 
x s + y % + 4s - 6y - 0. 

21. Intercept on x-axis equals 3, and passing through the points 
(-1,2) and (2, 3). 

22. Tangent to x-axis, radius 4, and abscissa of center 3. 

23. Tangent to y-axis, radius 2, and ordinate of center 4. 

24. Center on the line x — y + 2 = 0, and passing through the points 
(3, 7) and (1, 1). 

26. Center on the line 2x — y — 3 = 0, tangent to both axes, and in the 
first quadrant. 

26. Center on the line 2x — y — 3 = 0, tangent to both axes, and in the 
fourth quadrant. 

27. Center on the line 3x — y + 8 = 0, tangent to both axes, and in 
the. second quadrant. 

28. Radius 3, tangent to both axes, and in the second quadrant. 

29. Tangent to the line 3a; + y -f- 2 « at the point (-1, 1) and 
passing through the point (3, 5) 

30. Intercept on the y-axis 4, and tangent to the line x + 2y + 1 — 
at the point (—3, 1). 

31. Tangent to both axes, in the second quadrant, and also tangent 
to the line 3x — 4y + 30 — 0. (Two solutions.) 

32. Tangent to both axes, in the first quadrant, and also tangent to 
the line 3x - 4y + 30 = 0. 

33. Tangent to both axes and passing through the point (8, 1). (Two 
solutions.) 

34. Find the equation of the diameter with slope 2 of the circle 
x* - 4* + y* + 6y - 3 = 0. 

36. The point ( — 1, 2) bisects a chord of the circle x* + y* - 10. 
Find the equation and length of the chord. 

36. A chord of the circle x* + y* + 2x + 4y - 15 = is bisected by 
the point (—2, 1). Find the equation and length of the chord. 



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92 ANALYTIC GEOMETRY [§76 

37. Find the equation of the circle inscribed in the triangle whose 
sides are the lines 6a; + 7y « 86, — 7x + 6y « 86, and 2x — 9y — 86. 

88. Find the equation of the circle inscribed in the triangle whose 
sides are the lines Zx 4- 4y » 18, —4a; + 3y » 26, and y + 4 — 0. 

89. Find the equation of the circle circumscribing the triangle whose 
sides are the lines 7a; + 9y - 66, 3a; + V — 26, and x + 2y « 16. 

40. Prove analytically that an angle inscribed in a semicircle is a 
right angle. 

41. Prove analytically that a line from the center of a circle bisecting 
a chord is perpendicular to it. 

Suggestion. — Let the ends of the chord be (r, 0) and (6, c). 

42. Prove analytically that the length of a perpendicular from any 
point on the circumference of a circle to a diameter, is a mean propor- 
tional between the segments into which it divides the diameter. 

48. Prove that the length of the tangent from the point (xi, y\) to the 
circle x* + y* + Dx + Ey + F - is x x * + y x * + Dxi + Ey x + F - 0. 

76. Systems of circles. — If fi(x, y) = and f 2 (x, y) = 
are the equations of any two circles, then by article 67 
/i(s> y) + kf 2 (x, y) = is the equation of a curve through all 
the points of intersection of fi(x, y) = and f 2 (x, y) = 0. 
Furthermore in this case the curve will always be a circle or 
a straight line. 

To prove that this is so, let/i(», y) = stand for the equa- 
tion Aix 2 + Aiy 2 + DiX + 
Exy + Ft = 0, and let /,(*, y) 
= stand for A& 2 + A%y s + 
D2X + E 2 y + F % = 0. 

Then f 1 (x f y) + kf i (x f y) =0 
becomes A\x* + A x y % + D x x + 
Eiy + Ft + k(A&* + A t y* + 
D2X + E iy + F 2 )=0. 

Collecting like powers of x 
and y, this equation becomes 
(Xi + kA 2 )x* + (At + kA 2 )y* + (D x + kD t )x + (ft + *ft)y 
+ ft + Aft = 0. 

Since the coefficient of x 2 equals the coefficient of y 2 and the 
coefficient of xy equals 0, this is the equation of a circle. The 




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576] 



THE CIRCLE AND CERTAIN FORMS 



93 



exception occurs when Ai + kA* ■» 0, in which case, this 
equation is of the first degree and therefore is the equation of 
a straight line. 




Fio. 61. 

Example 1. — Find the equation of a circle through the point (1, 2) 
and the points of intersection of the circles 2s* + 2y* — 3s — 4y — 1 - 6 
and 3s* + 3y* - 8s - y - 4 - 0. 

Solution. — The equation of any circle through the points of intersection 
of these two circles is 

2s* + 2y* - 3s - 4y - 1 + fc(3s* + 3y* - Sx - y - 4) - 0. 

Since the point (1, 2) is on this circle its coordinates must satisfy the 
equation of the circle, therefore 

2+8-3-8-1+ &(3 + 12 -8-2- 4) =0. 

Solving for k, gives k « 2. 

Therefore the required equation is * 

2s* + 2y* - Zx - 4y - 1 + 2(3s* + 3y* - Sx - y - 4) - 0, 
or Sx* + Sy* - 19s - 6y - 9 - 0. 

Example 2. — Find the equation of the common chord of the circles, 
2s* + 2y* - 6s - 4y + 1 - and s* + y* - 2s - y + 3 - 0. 

Solution. — The equation of any circle through the points of inter- 
section of these two circles is 

2s* + 2y* - 6s - 4y + 1 + fc(s* + y* - 2s - y + 3) - 0. 

In order that this equation shall be the equation of a straight line, it is 
necessary that the coefficient of s* shall vanish, hence 2 + h = 0. This 
gives A; — —2 



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94 ANALYTIC GEOMETRY [§77 

Making this substitution the equation becomes 

2s* + 2y* - 6s - 4y + 1 - 2(s* + y* - 2s - y + 3) - 0, 
or 2s + 2y + 5 - 0. 

This is the equation of their common chord. 

If the two circles intersect in real points, the straight line 
thus obtained is their common chord, since it passes through 
their two points of intersection. If the two circles do not 
intersect visually, they are still said to intersect algebraically, 
their points of intersection being imaginary, and the line 
/i (z, y) + kfz (x, y) passes through their imaginary points 
of intersection. The straight line which passes through the 
real or imaginary points of intersection of two circles is called 
their radical axis. 

EXERCISES 

Find the equation of the common chord or the radical axis of the 
circles in exercises 1-6. 

l.x , + l/ , -3x+ y-6=0, 4. 2s* + 2y* - 3s - Zy + 5 - 0, 
s* + y* - 5s - Zy + 4 = 0. 3s* + Zy 1 - 2x - Zy + 4 - 0. 

2. s* + y* - 6s - Sy + 3 = 0, 6. 4s* + 4y* - x + y - 6 - 0, 
x s + y* + 4s + 2y - 7 - 0. 3s* + 3y* - 2x - 3y + 4 - 0. 

3. x* + y* - 3s - 4y + 2 - 0, 6. 3s* + 3y* - 2s - Zy + 6 - 0, 
s* + y* - 2s - 2y + 6 - 0. 2s* + 2y* + s + y - 2 = 0. 

7. Find the equation of the circle through the point (1,1) and through 
the points of intersection of the circles 

s* + y* - 2s - Zy + 4 - 0, 
s* + y* - 4s - by + 6 - 0. 

8. Find the equation of the circle through the point (3, 4) and through 
the points of intersection of the circles 

s* + y* - 7s - Zy + 10 - 0, 
s* + y* - 8s + 2y - 6=0. 

9. Prove that the common chords of the following circles, taken two at 
a time, meet in a point: 

s* + y* - 4s - Zy + 6 = 0, 
s* + y* - 2s + by - 2 - 0, 
s* + y* + s + 2y - 4 - 0. 

77. Locus problems involving circles. — Although the ele- 
ments dealt with in plane geometry are the point, straight 



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§77] 



THE CIRCLE AND CERTAIN FORMS 



95 



Y 


\ 




J>f f * ,lf) 




^)J 7a4 /r(24) I 


\ ' ° 


— i 1 i — 4**x 



line and circle, nevertheless the locus problems that can 
readily be handled by plane geometry are only of the simplest 
kind. On the other hand analytic geometry lends itself 
easily to the solution of locus 
problems as is illustrated by 
the following example. 

Example. — Find the locus of the 
point, which moves so that the sum 
of the squares of its distances from the 
points (0, 1) and (2, 1) is constant 
and equal to 20. 

Solution. — Let P(x, y) be any point ' 
on the locus, then 

FS* + FT* = 20, (1) 

FS* - x* + (y - l)*, 
FT 1 - (x - 2* + (y - 1)*. 
Substituting these values in equa- 
tion (1) 

x* + (y - 1)» + (x - 2)» + (y - 1)* 
Simplifying, a; 2 + y* — 2x — 2y = 7. 
Completing the squares in the x and y-terms, 

(x - 1)» + (y - 1)« - 3*. 
Hence the required locus is a circle whose center is the point (1, 1) 
and whose radius is 3. 

EXERCISES 

1. Find the locus of a point which moves so that the sum of the squares 
of its distances from (—2, 0) and (2, 0) is constant and equal to -26. 

2. Find the locus of a point which moves so that the sum of the squares 
of its distances from ( — 1, 2) and (2, 1) is constant and equal to 10. 

3. Find the locus of a point such that its distance from the point 
(—2, 0) shall always be twice its distance from the point (2, 0). 

4. Find the locus of a point moving so that its distance from the line 
Zx + 4y — 5 =0 shall equal the square of its distance from the point (1, 0). 

6. Find the locus of a point such that its distance from the y-axis shall 
equal the square of its distance from the point (0, 2). (Two solutions.) 

6. In an isosceles triangle of base 6 and equal sides of length 5, a point 
moves so that the product of its distances from the equal sides equals the 
square of its distance from the base. Prove one of the loci to be a circle 
and find its radius. 



Fig. 62. 



20. 



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96 ANALYTIC GEOMETRY [§78 

7. Find the locus of the vertex of a right angle if its two sides always 
pass through the points (—2, —4) and (2, 6). 

8. Find the locus of the vertex of an angle of 30°, whose sides pass 
through the points (—2, 0) and (2, 0). (Two solutions.) 

9. Find the locus of the vertex of a triangle, if the remaining two 
vertices are at the points (—3, 0) and (3, 0) and the length of the median 
from the vertex ( — 3, 0) is constant and equal to 5. 

10. The ends of a straight line of length 6 rest on the axes, find the 
locus of its middle point. 

78. Equation of a circle in polar coordinates. — Let the 
radius of the circle be r, and let C(pi, 0i) be the coordinates of 
its center, Fig. 63. 

Then if P(p, 0) is any point on the 
Jr^^l^ circle, by trigonometry, 

r 2 = p 2 + Pl 2 - 2p Pl cos COP. 
Replacing angle COP by its value 
(0 - 0i), 
r 2 = p 2 + Pl 2 - 2 PPl cos (0 - 0i). 
"*" The forms of this equation which 

Fiq. 63. occur most frequently are those where 

the center is the pole or where the 
circle passes through the pole and the center of the circle is 
either on the initial line or on the line = 90°. 

If the center is the pole, pi = 0, and the equation becomes 

p = r. 
If the circle passes through the pole and has its center on 
the initial line, 0i = and p\ = ±r. The equation of the 
circle then becomes 

p = 2r cos 0, or p = — 2r cos 0, 
according as the center is on the initial line or the initial line 
produced through the pole. 

If the circle passes through the pole and its center is on the 
line = 90°, B\ = 90° and p = ±r, and the equation becomes 

p = 2r sin 0, or p = — 2r sin 0, 
according as the circle lies above or below the polar axis. 




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§78] THE CIRCLE AND CERTAIN FORMS 97 

EXERCISES 

Find the equations of the following circles in polar coordinates: 

1. The center is at the pole and the radius equals 2. 

2. The center is at the point (5, 0) and the radius equals 5. 

3. The center is at the point (—4, 0) and the radius equals 4. 

4. The center is at the point (3, far) and the radius equals 3. 
6. The center is at the point (—2, far) and the radius equals 2. 

6. The circle is tangent to the initial line at the pole and the radius 
equals 6. 

7. The circle is tangent to the line $ » 90° at the pole and the radius 
equals 6. 

8. The center is at the point (3, 1*-) and the radius equals 3. 
Change from rectangular to polar coordinates. 

9. x* + #* - 6. 11. 2x* + 2y* + 5x - 0. 
10. x* +-y* - 3y - 0. 12. x* + y* - Ox - 8y - 0. 
Change from polar to rectangular coordinates and find the center and 

radius of each of the following circles. 

13. p + 6 sin $ = 0. 17. p + 2 cos $ + 3 sin $ - 0. 

14. p - 4 cos $ = 0. 18. p* + 3p cos $ + 4p sin $ - 6 - 0. 
16. p = cos $ + sin $. 19. p* = 9 sec*0 — p f tan*0. 

16. p = 5. 20. p f - 4 cscf - p* cot»0. 



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CHAPTER VI 



THE PARABOLA AND CERTAIN FORMS OF THE 
SECOND DEGREE EQUATION 

79. General statement. — It is an interesting and useful 
fact that an equation of the second degree in two variables, 
if plotted with reference to rectangular axes, gives a conic 
8ection y or simply a conic. That is, the graph is some plane 
section of a right circular cone. 

80. Conic sections. — When a plane intersects a circular 
cone there may be formed a circle, a parabola, an ellipse, an 

hyperbola, or, for cer- 
tain positions of the 
plane, a point, two in- 
tersecting straight lines, 
or two coincident lines. 
In Fig. 64, plane C is 
perpendicular to the axis 
of the cone and forms a 
circle; plane disinclined 
to the axis but intersects 
only one nappe of the 
cone and forms an 
ellipse; plane P is par- 
allel to an element of 
plane H intersects both 
The intersec- 




Fio. 64. 



the cone and forms a parabola; 

nappes of the cone and forms an hyperbola. 

tion is a point when a plane passes through the point V 

only; two intersecting straight lines are formed when the 

plane passes through V and intersects the nappes; and two 



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§81] THE PARABOLA AND CERTAIN FORMS 99 

coincident lines are formed when the plane passes through V 
and is tangent to the cone. 

The conic sections were first studied by the Greeks, who 
discovered and discussed their properties by methods of 
geometry. The modern method of studying these figures is 
by the help of algebra, which makes the treatment much 
simpler. For the purposes of this method of treatment, 
other definitions of the conic sections are given; but it can be 
readily shown that these definitions agree with the definitions 
mentioned above. 

EXERCISES 

1. Explain how a conic section could be two lines inclined to each 
other at an angle of 45°. Could the two straight lines formed on the 
same cone form different angles with each other? 

2. If the vertex angle of a cone is 30°, what would be the angle between 
the intersecting lines formed by the plane intersecting the cone? 

3. In forming an hyperbola, does the plane have to be parallel to the 
axis of the cone? Could hyperbolas of different shapes be formed on 
the same cone? 

4. Explain how a parabola of different widths could be formed on 
the same cone. 

5. Explain how ellipses of different widths could be formed on the 
same cone. Explain the change in the shape of the ellipse formed by a 
plane that revolved into a position parallel to an element of the cone. 

81. Conies. — A definition of a conic section, and one that 
can readily be translated into algebraic language, is the 
following: A conic is the locus of a point that moves in the 
plane of a fixed straight line and a fixed point not on the line, 
in such a manner that its distance from the fixed point is in a 
constant ratio to its distance from the fixed line. 

The fixed point is called the focus of the conic, and the 
fixed line is called the directrix. The constant ratio is called 
the eccentricity and is usually represented by e. 
. < The constant e is positive, and may be equal to 1, less than 1, 
or greater than 1. 



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N 
<-H p.v) 



N' 



V 2 pO 



100 ANALYTIC GEOMETRY [§82 

If e = 1, the conic is a parabola. 
If e < 1, the conic is an ellipse. 
If e > 1, the conic is an hyperbola. 

82. The equation of the parabola. — By the definition of 
the preceding article, the parabola is the locus of a point 
equidistant from the focus and the directrix. 

In Fig. 65, let F be the focus and D'D the directrix. Choose 

as x-axis the line X'X through 
F and perpendicular to D'D 
p(*.v) at R. The point on X'X 

midway between R and F is 
a point on the locus. Choose 
this point as origin. Then 
—*x Y'Y parallel to D'D is the 
y-axis. 

Let p represent the length 
and direction of RF. Then the 
coordinates of F are ($ p, 0), 
and the equation of D'D is 

x = -£p.. 

To derive the equation of the parabola, let P(x, y) be any 
point on the locus, and draw FP, and NP perpendicular to D'D. 

By definition FP = NP. 

But FP = V(x - £p) 2 + y\ and NP = x + \p. 

Then V(* - \pY + y 2 - * + ip. 

Squaring and simplifying, this becomes 
[28] iyj = 2px. N > 

The simple form of this^ equation is due to the choice of the 
coordinate axes. If they had been chosen differently, the 
equation would be more complicated; but the locus itself 
would be unaltered. 

Equation [28] is the required equation. For it has been 
proved true for every point on the parabola; and it is not true 



ViV FiViP.Q) 



Y c 

Fio. 65. 



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583] 



THE PARABOLA AND CERTAIN FORMS 



101 



for any point that is not on the parabola, for then FP is not 
equal to NP, and therefore y 2 is not equal to 2px. 

It should be remembered, that in the equation y 1 = 2px, 
p represents the length and direction of RF. Therefore, when 
the focus lies to the right of the directrix, p is positive; but, 
when the focus lies to the left of the directrix, p is negative. 

83. Shape of the parabola. — The shape of the parabola and 
its position relative to the coordinate axes can be readily 
determined from the equation y 2 = 2px. Solving for y gives 
y = ±\/2px. 

For any positive value of p we have: 

(1) When x = 0, y = 0. 
Hence the curve passes 
through the origin. 

(2) For all positive values 
of x, y has two numerically 
equal values but opposite in 
sign. Hence' the curve is 
symmetrical with respect to 
the x-axis. 

(3) For any negative value 
of x, y is imaginary. Hence 
nd part of the curve is at the 
left of the y-axis. As x in- 
creases from 0, the positive value of y increases and the 
negative value decreases. 

The curve can be located more precisely by the following 
points: 




Fig. 66. 



X 





\v 


P 


2p 


4p 


Sp 


50p 


y 





±p 


±py/2 


±2p 


±2pV2 


±4p 


±10p 



The parabola has the shape shown in Fig. 66. It is evident 
that all parabolas have the same shape, the appearance 



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102 



ANALYTIC GEOMETRY 



1184 



depending only upon the sue of the unit chosen. For a 
negative value of p, the parabola will be exactly the same 
shape but opening toward the left. 

84. Definitions, — The point of the parabola midway between 
the focus and the directrix is called the vertex of the 
parabola. 

The line through the focus and perpendicular to the directrix 
is called the axis of the parabola. As has been proved in the 
preceding article, the axis bisects all the chords of the parabola 
which are parallel to the directrix, once the axis of the para- 
bola lies on the £-axis. 

The chord of the parabola through the focus and perpen- 
dicular to the axis is called the latus rectum. The length of 





FiO. 67. 



»*• 2 PV , V positive x*m 2py . v negative 

Fio. 68. 



the latus rectum is the absolute value of 2p. For the ab- 
scissa of the focus is Jp, and, when « » |p, y =r ±p m 

In Fig. 67, V is the vertex of the parabola, VX is the axis, and P'P 
is the latus reetum. 

A parabola can be readily sketched if the position of the 
vertex V and focus F and the length of the latus rectum, 
P'P, are known. 

85. Parabola with axis on the y-axis. — The equation of 
a parabola whose axis is on the y-axis and whose vertex is at 
the origin is obviously obtained by interchanging x and y 
in the work of article 82. The equation is 



120] 



2py. 



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§861 THE PARABOLA AND CERTAIN FORMS 103 

The focus is at the point (0, Jp), and is on the positive or 
the negative half of the y-axis according as p is positive or 
negative. If p is positive, the parabola, Fig. 68, is above 
the x-axis; and, if negative, it is below the x-axis. 

It is to be remembered that the origin is at the vertex of 
every parabola whose equation is of the form [28] or [29]. 
These forms are called the standard forms of the equation of 
the parabola. 

EXERCISES 

1. Plot the following parabolas: y* = 2x t y* = — 2x t x* = 2y t and 

2. Give the coordinates of the foci of the parabolas in exercise 1. 
Give the equations of their directrices. What are their latera recta? 

3. Plot y* = 4x, using successively A hi., i in., J in., i in., 1 in., 
and 2 in. as a unit. 

4. Plot y* — ix using 4 in. as a unit. Plot y* « \x using 1 in. as a 
unit. Plot y* — x using i in. as a unit. Plot y* « 4x using J in. as a 
unit. Are all parabolas of the same shape? 

5. Write the equation of a parabola whose vertex is at the origin and 
focus at (1) (3, 0), (2) (0, 6), (3) (-4, 0), (4) (0, -2). 

6. Find the equations of the following parabolas, and give the latus 
rectum of each: 

(1) Vertex at origin, axis on x-axis, and passing through the point 
(2, 4). 

(2) Vertex at origin, axis on y-axis, and passing through the point 
(2, 4). 

7. The cables of a suspension bridge hang in the form of a parabola. 
Find the equation for such a cable in a bridge 1000 ft. between supports 
if the distance from the lowest point of the cable to the level of the top 
of the piers is 50 ft. 

Suggestion. — Take the origin at the lowest point of the cable. Then 
the point (500, 50) is on the parabola. Substitute these values in [29] 
and solve for p. 

8. Derive equation [29] from [28] by revolving the coordinate axes 
through an angle <p = —90°. 

86. Equation of parabola when axes are translated. — 

Transform the equation y 2 = 2px by translating the axes to 
anew origin at the point 0'(— h, —A), Fig. 69. 



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104 



ANALYTIC GEOMETRY 



[§86 



Y' 



->X 



By [12], x = x' — h and y = y' — fc. Substituting these 
values in y 2 = 2px gives 

(y / - fc ) 2 = 2p(x' - A). 
This is the equation of a parabola having its vertex at the 

point (h, k) when referred to the 
new coordinate axes, that is, the 
x*, y'-axes. If the primes are 
dropped, this becomes 
[30] (y - k)* = 2p(x - h), 
which is a convenient form for 
writing the equation of a para- 
bola with vertex at point (h, k) 
and axis parallel to the 3-axis. 
If p is positive, the parabola opens toward the right; and if 
negative, it opens toward the left. 

Similarly, when the axis of the parabola is parallel to th£ 
y-axis, the equation is 
[30x] (x - h) 2 = 2p(y - k). 



(-A.-*> 



-+X' 



Fio. 69. 




*»X 



(1) \^ (2) 

p positive p negative 



(3) / (1) 

p pbsitive p negative 

Fio. 70. 



The position of these parabolas with reference to the 
coordinate axes is shown in Fig. 70. 

Example 1. — Find the equation of a parabola with vertex at the point 
(2, —3), axis parallel to the a>axis, and p = 2. Plot. 

Substituting in [30], (y + 3) 2 = 2 X 2(x - 2). 

Simplifying, y* + 6y - 4s + 17 = 0. 

The curve is plotted in Fig. 71. 



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§861 THE PARABOLA AND CERTAIN FORMS 105 

Example 2. — Fine} the equation of the parabola whose vertex is at 
the point (3, —6), axis parallel to the y-axis, and which passes through 
the point (-3, -10). 

Solution. — The equation is of the form Y 

[30i], in which h — 3, A; — —6, and p is 
to be found. x 

Substituting in [30i], 

(_3 - 3)*«2p(-10, +6). 

Solving for p, p — —4$. 

Substituting values of h, k, and p in 
[80i] gives 

(*-3)*«2(-4i)(y + 6). 

Simplifying, x* - 6x + 9y + 63 = 0, r' 

the required equation. Fia. 71. 

EXERCISES 

1. Write the equations of the following parabolas: 

(1) Vertex at (3, 4), p =4, and axis parallel to the re-axis. 

(2) Vertex at (2, 3), p = —4, and axis parallel to the x-axis. 

(3) Vertex at (— 6, 2), p «■ 6, and axis parallel to the s/-axis. 

(4) Vertex at (2, — 3), p = —3, and axis parallel to the y-axis. 

2. In each part of exercise 1, give the codrdinates of the focus, equa- 
tion of the directrix, and plot the parabola. 

3. Write the equations of the parabolas with vertex of each at 
(—4, —2), latus rectum of each equal to 10, and axes parallel to x-axis. 

4. Write the equation of the parabola with vertex at (3, —2), origin 
on the directrix, and axis parallel to y-axis. 

5. Transform x* + By = 12 to new axes parallel t* the old, with the 
new origin at the point (2, 5). 

6. Find the equations of the following parabolas, and sketch each 
curve: 

(1) Vertex at (4, 5) and the focus at (6, 5). 

(2) Vertex at (-4, 2) and the focus at (-4, 4). 

(3) Vertex at (-4, 2) and the focus at (-6, 2). 

(4) Vertex at (3, -4) and the focus at (3, -6). 

7. Find the equations of each of the following parabolas, and sketch 
each curve: 

(1) Vertex at (2, 3), axis parallel to x-axis, and passing through the 
point (5/6). 

(2) Vertex at (3, —2), axis parallel to x-axis, and passing through the 
point (-1, 3). 



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106 ANALYTIC GEOMETRY [§87 

I 

(3) Vertex at (2, 3), axis parallel to y-axis, and passing through the 
point ( — 1, 1). 

(4) Vertex at (3, —2), axis parallel to y-axis, and passing through the 
point (-1, 3). 

87. Equations of # forms y 2 + Dx + Ey + F = <ah!i 

x 2 + Dx + Ey + F = *0. — (1) Every equation of the form 
y 2 + Dx + Ey + F = 0, where D j* 0, represents a parabola 
whose axis is parallel to the x-axis. 

(2) Every equation of the form x 2 + Dx + Ey + F = 0, 
where E j* 0, represents a parabola whose axis is parallel to 
the y-axis. 

Proof of (1).— Given y 2 + Dx + Ey + F = 0, where Z)>* 0. 

E 2 E 2 -rH 

Completing square in y, y 2 + Ey + -r- = — Dx + -j- — F. 



Or 



/ , E\ 2 n/ E 2 -4F\ •'••'♦ 

(y + 2) - - D (* --&-)• 



E 2 — 4F E 

This is in the form of [30], where h = — j~ — , k = — ^ 

A D U 

and p « -^ 

Therefore the equation j/ 2 + Z)s + Ey + F = where D j* 
represents a parabola whose axis is parallel to the x-axis. 
The proof of (2) is similar to that of (1). 

Example 1. — Transform the equation 

y* + 4y — 4s + 8 =0 into the form of 

[30], give the coordinates of the vertex 

x and focus, write the equations of the axis 

and directrix, and sketch the parabola. 

Solution.-— Completing the square in y, 
y* + 4y + 4 = 4x - 8 + 4. 
Or (y + 2)» - 4(s - 1). 
Hence the vertex is at the point 
V (1 - 2). 
Since 2p = 4, p = 2, and the focus is 
Fio. 72. one unit to the right of the vertex, or at 

the point (2, -2). 
The axis is parallel to the x-axis and two units below. Hence its 
equation is y = — 2* 



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§88] THE PARABOLA AND CERTAIN FORMS 107 

The directrix is perpendicular to the x-axis and one unit to the left 
of the vertex. Hence its equation is x - 0. 

The parabola is shown in Fig. 72. 

Example 2. — Find the equation of the parabola with its axis parallel 
to the x-axis, which passes through the points (0, 1), (2, 3), (5, 2). 

Solution. — The equation is of the form y* + Dx + Ey + F - 0. 

Since the parabola passes through the point (0, 1), these coordinates 
satisfy the equation. Substituting these coordinates gives 

1 + E + F - 0. 

Likewise (2, 3) give 9 + 2D + SE + F - 0. 
And (5, 2) give 

4+ 52> + 2# + F « 0. 

Solving these equations for 2>, E, and F, D - J, E - — ¥, and F - *f. 
Substituting these values in y» + Dx + Ey + F - 0, gives 

y 1 + \x - V* + ¥ - 0. 
Or 4y* + x — 17y + 13 — 0, the required equation. 

, 88. The quadratic function ax 2 + bx + c. — The locus of 
the equation y = ax 2 + bx + c, where a, 6, and c are real 
numbers and a 5*0, is a parabola with axis parallel to the 
y-axis. 
To see this, reduce the equation to the standard form [30J. 

b 2 — 4oc 



/ b\ 2 

Completing the square in x> y = o ( x + =- ) 4 

-_4oc\ 
4a / 



/ « & \ 2 1 / , &* - 4ac \ 



This is in the form (x — h) 2 = 2p(y — fc), where h - — =- 
and fc = j > and is a parabola with vertex at the 

point (— «- * Z ) anc * ax ^ on ***© ** ne x "t" 9~ "" ®' 

Evidently the parabola opens upward if a > and downward 
ifa<0. 

89. Equation simplified by translation of coordinate 
axes. — It is evident that y 2 + Dx + Ey + F = and 
x 2 + Dx + Ey + F = can be transformed to the forms of 



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108 



ANALYTIC GEOMETRY 



I§89 



[28] and [29], respectively, by a suitable translation of the 
coordinate axes. 

. In the first equation, the term in y and the constant term 
can be made to vanish; and, in the second, the term in x and 
the constant can be made to vanish. 

Example. — Translate the coordinate axes so as to transform the 
equation y* + 6s — 4y + 10 = to the form of y* = 2px. 

Solution. — Using [12], put x — z? + h and y — y' + k, then 
(y' + *)» + 6(x' + h) - My* + *) + 10 = 0. 

Or y 7 * + 6V + (2fc - 4)y> + (A; 1 - 4* + 6h + 10) - 0. 

In order that the y' term and the con- 
stant term shall vanish 
2fc - 4 = and h* - 4fc + 6h + 10 - 0. 

Solving these equations, h = — 1 and fc = 2. 

Therefore the transformed equation is 
y** - -6*'. 

The transformation can also be made by 
completing the square in y, whence 
y* - 4y + 4 = -6s — 6, 
or (y - 2)» = -6(s + 1). 

Put y — 2 = y' and 3 + 1 - s', and 
obtain y' 1 = — 6s', as before. 

The curve is plotted in Fig. 73. 




*~x 



+-X 



Fio. 73. 



EXERCISES 

1. Transform the equations of the following parabolas to the form 
of [30] or [30 J; and in each case give the coordinates of the vertex and 
the focus, write the equations of the axis and directrix, and plot. 

(1) y* - 4x - 4y + 16 ~ 0. (4) 2x* - 24x + 3y + 78 = 0. 

(2) y 1 + 2x + 8y + 6 = 0. (5) 3y» + lfo' - 12y + 20 = 0. 

(3) 4x* + \2x - 20y + 49 = 0.- (6) 2x* - 18s + 15y - 21 - 0. 

2. Find tfce equation of the parabola with axis parallel to the 
y-axis, which passes through the points (2, 3), (1, 0), and (0, 2). 
Find the coordinates of the focus and vertex of this parabola, and its 
latus rectum. 

3. Find the equation of the parabola which has the line y = 4 as axis, 
the line x = — 2 as directrix, and p = 6. 

4. Find the equation of the parabola which has its vertex at (2,-3), 
its axis parallel to the z-axis, and which passes through the point (5, 2). 



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§90] THE PARABOLA AND CERTAIN FORMS 109 

5. Translate the coordinate axes so as to transform the following 
parabolas to the form of [28] or [29]. In each case plot showing both 
sets of axes. 

(1) y* - 4x - 6y + 8 - 0. (3) y* + 8x - 4y - 4 - 0. 

(2) x* - 8x + 16y - 0. (4) 3x» + 6x - 7y + 8 - 0. 

6. For each of the parabolas of exercise 5, find the equation of the 
directrix with reference to both sets of axes. Give the coordinates of 
the focus for both sets of axes, and the value of the latus rectum. 

7. Plot the equation y — ax 1 + bx + c discussed in article 88 for 
(1) &» - 4oc > 0, (2) 6* - 4ac « 0, (3) 6* - 4ac < 0, both when 
a > and when a < 0. 

90. Equation of a parabola when the coordinate axes are 
rotated. — Transform the equation y 2 + Dx + Ey + F = by 
rotating the coordinate axes through an angle <p, using the 
formulas [13]. 

Putting x = x' cos <p — y' sin <p, and y = x f sin <p + y 9 cos <p, 
in y 2 + Dx + Ey +F = 0, gives (a/ sin ¥> + !/' cos *>) 2 
+ Z)(a/ cos <p — j/' sin ^) + E(x' sin ¥> + y' cos <p) + F = 0. 

Collecting terms, 

x' 2 sin 2 ip + 2 sin ¥> cos <p x'y' + y' 2 cos 2 <p 
+ (D coa <p + E sin *>)z' + (E cos <p-D sin *>)y'+F = 0. (I) 

A similar form is obtained from x 2 + Dx + Ey + F = 0. 

If the angle of rotation is some multiple of 90°, then 
2 sin <p cos <p = 0, and the coefficient of x'y' is 0. Hence, 
in this case, the x'y'-tevm vanishes. 

If the coordinate axes are rotated through an angle <p f such 
that the axis of a parabola is not parallel to either coordinate 
axis, the equation of a parabola is of the form 

Ax 2 + Bxy + Cy 2 + Dx + Ey+F = 0, (II) 

the most general form of an equation of the second degree 
in x and y. » 

It is readily seen that in equation (I), B 2 — 4AC = 0. It 
will be shown later, Art. 121, that the necessary and sufficient 
condition that any equation of the form of (II) represents a 
parabola is that B 2 — 4AC = 0. 



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110 



ANALYTIC GEOMETRY 



[§90 



Example 1. — Transform the equation x % + 2x — 3y + 4 — by 
rotating the coordinate axes through an angle of 45?. Plot. 

Solution. — Substituting x — x' cos 45° — y' sin 45° 
and y — x 9 sin 45° + j/ 7 cos 45°, 

(a/ cos 45° - y' sin 45°)* + 2(x / cos 45° - tf sin 45°) 
- 3(0/ sin 45° + y> cos 45°) +4=0. 
Simplifying, *>* - 2*V + y 71 - V2*' - 5 W +8 = 0. 
Example 2. — By rotating the coordinate axes transform the equation 
9x* -^i^+ 16V — 116x - 162y + 221 = 0, to a form which contains 
no term m xy. 
Solution. — Putting x = a/ cos *> — y* sin ?, and y = a/ sin <p + y* cos <p, 
9(x / cos *> — y' sin ^) s — 24(s' cos tp — y' sin ^Xs' sin ? + y' cos *>) + 
16(3' sin *> + y 7 cos *>)* — 116(:r/ cos tp — y' sin *>) — 
162(x / sin ? + y> cos <p) + 221 = 0. 
Collecting terms, (9 cos 1 *>— 24 sin ? cos *> + 16 sin 1 <p)x* % + • 

(14 sin ^> cos ^> + 24sin 1 *> — 24cos , *>)x / y / + 
(9sin , ^ + 24sinv>coS^ + 16 cos 1 *>) y' 1 — 
(162 sin *> + 116 cos <p)x' + * 

(116 sin *>- 162 cos ^Jy 7 + 221 = 0. 
Now, in order that the x*\f term shall 
vanish, its coefficient must be 0. Hence 
^24 sin 1 y — 24 cos 1 <p + 14 sin ^ cos y = fl. 
Or -24 cos 2tp + 7 sin 2? - 0. 

Dividing by cos 2?, 7 tan 2*> = 24, or 
tan 2<p - *#>. 

From this by trigonometry, cos 2? = /j. 




Then 



And 



— cos 



2 



COS <p 



+ cos 2^ 






"2~ *' 



+ A 



f 



Substituting these values for sin <p and cos *> in the above equation 
and simplifying, 25^ 1 _ = ^190aj / - 60y' + 221 - 0. 

EXERCISES 



1. Transform the equations y 1 = 2px and x 1 = 2py by rotating the 
coordinate axes through an angle of 90°. 

2. Transform the following equations by rotating the coordinate 
axes through the angle given in each case: 

(1) y 1 - Ax. * = 46°. 



(2) a 1 + 3s - 2y + 6 - 0. 



= 30°. 



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§91] 



THE PARABOLA AND CERTAIN FORMS 



111 



(3) 4x* - 4xy + y* + 2x - 6y - 10 - 0. ? - sin" 1 iVS. 

(4) 9s* + 12xy + 4y« + lOx - 64y - 68 - 0. 2* - tan" 1 -^. 

3. Derive the equation of the parabola whose directrix is the line 
4x +3y + 2 — 0, and whose focus is at the point (2, 3). 

4. Simplify the following equations, and plot. First rotate the coordi- 
nate axes to free of xy-term, then translate to change to the standard form. 

(1) s 1 - 2xy +ty -6x-6y+9»0. 

(2) 2x» + Sxy + 8y* + x + y + 3 = 0. 

(3) x* + 2xy + y* - 12x + 2y - 3 - 0. 

91. Equation of parabola in polar coordinates. — Starting 
with the definition of article 81, the 
equation of parabola in polar co- 
ordinates can be easily derived. 

In Fig. 75, let be the fixed point 
(focus), and D'D the fixed line (direc- 
trix). Choose as pole and OX, 
perpendicular to D'D, as the polar 
axis. Let P(p, 6) be any point on the 
locus. Draw MP and NP perpen- 
dicular to OX and D'D respectively. 

By definition, OP = NP. 

But OP = p, and NP = QM - p + p cos 6. 

Hence p = p + p cos 0. 

Solving for p, 

This is the polar equation of a parabola referred to its focus 
and axis. 

EXERCISES 




**X 



Fio. 75. 



1. Given the equation p 



-, transform it to rectangular 



1 - cos $ 
coordinates and by translation of axes derive the equation y* — 2px. 

2. By taking the focus at the left of the directrix, derive the equation 

of the parabola in the form p = - — ■ • 

1 + COS0 

3. Change the following equation into polar coordinates with the 



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112 



ANALYTIC GEOMETRY 



[§92 



pole at the origin, and the polar axis on the positive part of the x-axis: 

y* - 2px + p*. 

4. Show that if the vertex of the parabola is taken as pole and the axis 

of the parabola as polar axis, the equation of the parabola in polar 

a : 2p cos 

coordinates is p = — r- = — • » 

sin 2 B 

92. Construction of a parabola. — First 
method. — The directrix D'D and the 
focus F are supposed known. 

Place a right triangle, Fig. 76, with 
one side CB on the directrix as shown. 
Fasten one end of a string whose length 
is CA, at the focus F and the other at 
A. With a pencil at P, keep the string 
taut and move the triangle along the directrix, 
and the point P will generate a parabola, 




MT 



Fio. 76. 



Then FP = CP, 
Why? 
Second method. — As before, 
the directrix D'D and the 
focus F are supposed known. 
In Fig. 77, draw MX 
through F and perpendicular 
to D'D. Draw any number 
of lines A' A, B'B, etc., par- 
allel to the directrix, and 
intersecting MX in Mi, M 2f 
etc. With F as center and a 
radius equal to MMi, strike 
arcs intersecting A' A in Pi 
and Qi. In like manner, with 
MM 2 as a radius, strike arcs 
intersecting B'B. Continue 
in like manner for the other 
lines drawn. Then the points, thus determined, lie on the 
parabola. Why? In this way the parabola can be located 
as accurately as desired. 



D 


A 


B 


C 
*P> 


F 


Pi 






MiP 


M % 


M* 


^4 


Mt 






Pi 


P» 










« 


A' 


B' 


c' . 




& 





Fio. 77. 



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§93] 



THE PARABOLA AND CERTAIN FORMS 



11 3» 



EXERCISES 

1. Construct a parabola by. the second method, in which p = 1 in. 
In which p = $ in. 

2. Construct a circle of radius 8 in., and a parabola with its vertex at 
the center of the circle, and its focus on the positive x-axis at the point 
midway between the center and 

circumference. Write the equa- 
tion of each in the standard 
forms, and compute the coordi- , 
nates of the points of inter- 
section of the curves. 

3. Explain how the con- 
struction shown in Fig. 78, 
determines a parabola. 

APPLICATIONS 

93. Parabolic arch. — 

The cable of a suspension p^ 78 

bridge hangs in the form 

of an inverted parabolic arch. Arches for bridges, when the 
weight is uniformly distributed, are properly constructed in 
the form of a parabola. In metal-arch bridges the loading is 
practically uniform on the horizontal, and so such bridge 
structures are in the form of parabolic arches. The arches 

of concrete bridges 
jjM are seldom if ever 

built in the form of a 
parabola, for, in such 
structures, the load- 
ing cannot be uni- 
formly distributed on 
the horizontal. 

In the parabolic 

arch, Fig. 79, AB = 2s is the span, and CO = h is the height. 

If the origin is taken at the vertex of the parabola, and the 

axis along the y-axis, the equation is of the form x 2 = 2py. 

To find the value of p, we know that the point B(s, —A) is 




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114 ANALYTIC GEOMETRY [§94 

on the parabola. Substituting these coordinates in x 2 = 2py, 

s 2 
gives s 2 = —2ph, and p = — st* Hence the equation of 

s 2 hx 2 
the parabola is x 2 = — j-y, and from this y = 2 * 

The height of the arch at any distance x from the center is 
NP = NM + MP = h + y - A - ~ 

EXERCISES 

1. A parabolic arch has a span of 120 ft. and a height of 25 ft. Derive 
the equation of the parabola, and compute the heights of the arch at 
points 10 ft., 20 ft., and 40 ft. from the center. 

2. A parabolic arch has a span of 40 ft. and a height of 15 ft. Find 
the height of the arch at intervals of 5 ft. from the center. 

3. The distance between the supports on the river span of the Brooklyn 
suspension bridge is about 1600 ft., and the vertex of the curve of the 
cables is 140 ft. below the suspension points. Find the equation of the 
curve if the lowest point is taken as origin. 

4. The towers supporting a suspension bridge are 320 ft. apart and rise 
80 ft. above the roadbed. The lowest point of the parabola formed by 
the cables is 20 ft. above the roadbed. Find the equation of the curve 
of the cables using as origin the point in the roadbed below the vertex of 
the parabola. 

94. The path of a projectile. — A projectile starting at the 

origin, Fig. 80, with an initial 
velocity of v ft. per second, 
and making an angle a with 
\ the horizontal, would after t 

\ * seconds have the position x 
B = v cos at and y = v sin at, 
if the action of gravity and 
the resistance of the air were 
not considered. If the action of gravity is considered, y is 
decreased by \gt 2 ft. in t seconds. Then the coordinates of 
the projectile at time t are 

x = v cos a % and y = v sin at — \gt 2 . (I) 



Fig. 80. 



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§94] THE PARABOLA AND CERTAIN FORMS 115 

The equation of the path of the projectile in rectangular 
coordinates is found by eliminating t between these equations 
and is 

^"^^-srib^* 1 - (II) - 

EXERCISES 

1. Eliminate t between the equations (I) and derive equation (II). 

2. Show that equation (II) is a parabola with its vertex at the point 



c- 



2 sin 2a v 2 sin 2 a\ , v 2 cos 2 < 

and p= — 



2(7 2g ) y g 

3. Find the s-intercept of (II), and thus find the range on the horizontal 

A . v 2 sin 2a 

to be 

9 

4. Find the height of the projectile when at a horizontal distance 
equal to one-fourth the range. 

5. Find the horizontal range when v = 2000 ft. per second and (1) 
a - 45°, (2) a • - 30°, (3) a = 60°. Use g = 32. 

6. Show that a projectile with a given velocity and at an angle of 
60°, rises three times as high as it would if the angle were 30°. 

7. What must be the initial velocity v of a projectile, if with an 
angle of elevation of 20°, it is to strike an object 80 ft. above the horizon- 
tal plane of the starting point, and at a horizontal distance of 1000 yd. ? 

GENERAL EXERCISES 

1. The formula for the height of a bullet shot vertically upward 
with a velocity of 2000 ft. per second is s = 2000J - 16* 2 . Find the 
coordinates of the vertex, and plot the curve from which the height s 
at any time t may be read. 

2. When one variable varies directly as the square of another, the 
equation connecting the two variables will represent a parabola. The 
length of a pendulum varies as the square of the time of a beat. This 

IT 2 

gives the formula t 2 = -J, where t is time in seconds, g is 32, and I is 

y 
length in feet. Plot a curve from which can be read the time of a beat for 

lengths up to 20 ft. 

3. In a parabolic reflector, such as used for an automobile headlight, 
the source of light is placed at the focus of the parabola that is a section 
of the reflector. Find the position of the source of light in a reflector 
10 in. in diameter and 5 in. deep. 



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116 ANALYTIC GEOMETRY [§94 

4. Find the coordinates of the points of intersection of the parabola 
x* - Sy and the line 3s — 2y - 8 = 0. 

5. Find the equation of the straight line passing through the focus 
of the parabola y* = &x and making an angle of 45° with the axis of the 
parabola. 

• 6. What value must be given to A; if the line 3x+2y + Aj = 0isto 
be tangent to the parabola x* =* — 6y? Plot. 

Suggestion. — Eliminate y between the two equations. Since a tangent 
meets the curve in two coincident points, the two values of x in the 
resulting equation must be equal. Hence put the discriminant of this 
quadratic equation equal to zero and solve for values of k. 

7. Find the points of intersection of the following curves: x — Zy = 
and y* - 3s - Qy + 14 = 0. 

8. For what values of m is the straight line y = mx + 2 tangent to 
the parabola x* - 6x + Sy + 41 = 0? 

9. One end of a chord through the focus of a parabola is at the point 
(10, 10). Find the coordinates of the other end if the parabola has its 
vertex at the origin and its axis on the positive part of the rr-axis. 

10. Transform the following equations in polar coordinates into 
rectangular coordinates and simplify: 

(1) " = 1 + cos 6 (2) " = 5 - 5 cos (3) " " nT'coTtf" 

11. Plot the following curves given in polar coordinates and find the 
coordinates of their points of intersection: 

(1) p cos B - 4, p = -. — — -• (2) p - 4, p = 



1 - cos v ' * ' H 1 + cos 

12. Show that the equation p = 8 sec* $0 is that of a parabola, and 
sketch the curve. 

13. Find the equation of the circle circumscribing the segment of 
the parabola y* = 2px, cut off by the latus rectum. 

14. An equilateral triangle having one vertex at the origin is inscribed 
in the parabola y* — 2px. Find the length of a side of the triangle. 

15. Show that x* + V* = a* is the equation of a parabola. Sketch 
the curve. 

16. Find the equation of the parabola with x + y ™ as directrix, and 
focus at (ia, ia). Express in the form given in the previous exercise. 



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CHAPTER VII 

THE ELLIPSE AND CERTAIN FORMS OF THE SECOND 
DEGREE EQUATION 

. 95. The equation of the ellipse. — By the definition of 
article 81, the ellipse is the locus of a point whose distance 
from a fixed point, the focus, is to its distance from a fixed 
straight line, the directrix, in a constant ratio e, less than 1. 
In Fig. 81, let F be the focus and D'D the directrix. Choose 
as x-axis the line X'X through F and perpendicular to D'D 
at R. 




Fig. 81. 

Since e<l, there are two points V and V on X'X such that 

VF FV 

~y = e and ^yr = e « Hence the points V and V are on 

the locus. 

Choose 0, the point midway between V and F', as origin, 
and Y'Y through 0, parallel to D'D, as y-axis. 

Let the length of W = 2a. 

Then VO = OV - a. 

117 



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118 ANALYTIC GEOMETRY [§95 

It is necessary first to find the equation of the directrix and 
the coordinates of the focus. 

From the definition of the ellipse, 

VF = e RV, or a - FO - e(RO - a), (1)' 

and FV = eRV\ or a + FO = e(RO + a). (2) 

Adding equations (1) and (2), 

2a = 2eR0, or 720 = -• 

e , . 

Then the equation of the directrix is x = — • r 

Subtracting equation (1) from equation (2), 

2F0 = 2ae, or FO = ae. 
Then the coordinates of the focus F are ( — ae, 0). 
Now to derive the equation, let P(z, y) be any point on the 
locus* and draw FP, and NP perpendicular to D'D. 
By definition, FP = ei\TP. 

But FP - Vte + ae) 2 + y 2 , and iVP - ? + *. 

Then V(x + ae) 2 + y* = e(^ + x) . 
Squaring and arranging, this becomes 

a 2 ^ a 2 (l - e 2 ) ' 

Since e < 1, a 2 (l — e 2 ) is positive and less than a 2 . Let it 
be represented by b 2 and the equation of the ellipse is 

x 2 v 2 

This is a standard form of the equation of the ellipse, and is . 

the form in which the equation of the ellipse is usually written. 

Its simple form is due to the choice of the coordinate axes. A 

different choice of axes would give a less simple form of the 

equation, but the locus itself would be unaltered. 

Va 2 — b 2 

Since b 2 - a 2 (l - e 2 ), e - -^ — 

7 a 

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§96] 



THE ELLIPSE AND CERTAIN FORMS 



119 



. Equation [32] is the required equation of the ellipse. For it 

has been proved true for every point on the ellipse, and it 

can be readily proved that it is not true for any point that is 

not on the locus. The proof of this is left as an exercise. 

96. Shape of the Ellipse. — The shape of the ellipse and its 

position relative to the coordinate axes can be readily 

x 2 y 2 
determined from the equation — 2 + ^ = 1. 



Solving for x, x = iiW-?- 



Solving for.y, y = ± - Va 2 — x 2 . 

(1) For all values of y such that b 2 — y % > 0, x has two real 
values, numerically equal but opposite in sign. Wheny 2 = b 2 , 
x = 0. For all values of x such that a 2 — x 2 > p, y has two 
real values, numerically equal but opposite in sign. When 
x 2 = a 2 ,y = 0. Hence the curve is symmetrical with respect 
to both coordinate axes and the origin, and its intercepts are 
a and —a on the rc-axis, and b and —6 on the y-axis. 

(2) For all values of y such that b 2 — y 2 <0, x is imaginary; 
and for all values of x such y 
that a 2 — x 2 <0, y is im- 
aginary. Hence no part of 
the curve lies outside of the 
rectangle bounded by the 
four lines x = ±a and 
y = ±6. 

(3) As x increases from 
—a to 0, the positive value 
of y increases from to 6, 
and the negative value of 

y decreases from to —6. As x increases from to a, the 
positive value of y decreases from b to 0, and the negative 
value of y increases from —6 to 0. 
The ellipse has the shape shown in Fig. 82. 





CO. 6) 


B 


V-6 






1 

II 
a 

V 




^ 


9 
It 
H 

v 


(-«.0) 


F O 




(a,0 




(0.-6) 


B' 


»=-& 





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120 



ANALYTIC GEOMETRY 



.[§97 



The formula ft 2 = a 2 (l — e 2 ) can now be readily interpreted 
geometrically. For in the right triangle FOB, Fig. 82, FO= ae 
and OB = 6. ___ 

Then FB 2 = (ae) 2 +6 2 . 

But from ¥ - a 2 (l - e 2 ), a 2 - (ae) 2 + 6 2 . 



Hence 



a 2 = FB 2 , or a = FB. 




>-x 



97. Definitions. — The center of symmetry of the ellipse is 
called the center of the ellipse. 

The chord through the focus and center of an ellipse is called 

the major axis. Its length is 2a. 
One-half of the major axis is 
called the semimajor axis. 

The chord through the center 
of the ellipse and perpendicular 
to the major axis is called the 
minor axis. Its length is 26. 
One-half of the minor axis is 
called the semiminor axis. 

The chord of the ellipse through the focus and perpendicular 

26 2 
to the major axis is called the latus rectum. Its length is — t 

for the abscissa of the focus is — ae, and when x = — ae, 

The points on the ellipse at the ends of the major axis are 
the vertices of the ellipse. 

In Fig. 83, V'V is the major axis, B'B the minor axis, and P'P the 
latus rectum. 

An ellipse can be readily sketched if the position and lengths 
of the axes are known. 

98. Second focus and second directrix. Theorem. — An 
ellipse has two foci and two directrices. 

In Fig. 84, on OV take OF' = FO and OR' = RO. Draw 
E'E parallel to D'D. Then F' is also a focus and E'E the 
corresponding directrix of the ellipse. 



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§99] 



THE ELLIPSE AND CERTAIN FORMS 



121 



Proof. — Let P be any point of the ellipse. Thrbugh P 
draw PN parallel to the rc-axis and intersecting D'D in N. 
Because of the symmetry of the ellipse, PN intersects the 
ellipse at a second point P' and the line E'E at N'. Draw PF 
and P'F'. 

From the symmetry of the figure, FP « F'P', and NP = P'N'. 

F'P' 



FP 
But NP = € ' 



P'N' 



= e. 



Then the ellipse is also the locus of a point P' whose distance 
from F' divided by its distance from E'E is e. 

Therefore F' is a focus and E'E is the corresponding 
directrix of the ellipse, and the ellipse has two foci and two 
directrices. 

The coordinates of the foci are (±oe, 0), and the equations 

of the directrices are x = +— 




Fig. 84. 



Fig. 85. 



99. Ellipse with major axis on the y-axis. — The equation 
of an ellipse whose major axis is on the t/-axis, and whose 
center is at the origin is obviously obtained by interchanging 
x and y in the work of article. 95. The equation then is 

V 2 X 2 

[33] ?i + £-,-l. 



Here the major axis is 2a as before; the minor axis is 26; the 
coordinates of the vertices are (0, ±a); the coordinates of 



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122 ANALYTIC GEOMETRY [§99 

the foci are (0, ±ae); and the equations of the directrices are 

EXERCISES 

1. In each of the following ellipses find the semimajor axis, the semi- 
minor axis, the eccentricity, the coordinates of the foci, and the equa- 
tions of the directrices. Sketch each ellipse. 

(3) 4x* + 9y* - 36. (6) 6s* + 9y* - 54. 

2. Find the distance from the foci to the ends of the minor axis in 

x 1 y 1 
the ellipse — „ + tz = 1. 
a 2 o z 

3. Write the equation of an ellipse with center at the origin, and major 
axis on the x-axis, having given: 

(1) o - 6, b - 4. (4) Focus at (5, 0), e - J. 

(2) a = 4, c = J\/3. (5) Directrix is x =7, e = }. 

(3) 6 = 3, e = J. (6) Latus rectum = 4, a = 8. 

(7) Focus at (\/3, 0), directrix is x = 3-\/3. 

£ 2 v 2 

4. In the ellipse — + — = 1, find the values of y when x = 2, when 

25 16 

x «■ 4, when £ = 5, when x = 6. 

x 2 y* 

5. Find the length of the latus rectum in the ellipse — + r- «■ 1. 

x* y* 
In the ellipse — + — = 1. 

6. Find the equation of the ellipse with center at the origin, axes 
on the coordinate axes, and passing through the points (1, f\/3) and 
(K/6, 1). 

7. Derive equation [33] from [32] by rotating the coordinate axes 
through an angle <p = 90°. 

8. Find the semi-axes, eccentricity, and the latus rectum of each 
of the following ellipses: 

(1) 6y* - 30 - 5s*. 

(2) 2x* + y* - 2m, ro>0. 

(3) x* + qy* - «, g>l, and s>0. 

(4) px* + gy* = pg, p>0, g>0, and g>p. 



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§100] 



THE ELLIPSE AND CERTAIN FORMS 



123 



9. Find the distances from the foci of the ellipse — + •— = 1 to a 

point on the ellipse, whose abscissa is 2. 

10. The minor axis of an ellipse is 24, and the foci and origin divide 
the major axis into four equal parts. Find the equation of the ellipse. 

11. Assume the equation of the ellipse, — + r. = h BJi ^ show that 

the sum of the distances of any point on it from its foci is 2a. 

12. Regard the circle as an ellipse with a — b f and find its foci, direc- 
trices, and eccentricity. 

13. Find the coordinates of the points of intersection of the ellipse 
2x* + Zy* = 14 and the parabola y* — 4x. 

14. Find the locus of the vertex of a triangle if the base is 2a, and 

b* 
the product of the tangents of the angles at the base is — 

c* 

Suggestion. — Take the x-axis on the base and the origin at the center. 

15. Find the locus of the vertex of a triangle if its base is 26 and the 
sum of the other sides is 2a. Take the x-axis on the base and the origin 
at the midpoint. 

x^ t/' x' t/* 

16. Discuss the equations — -f ti = 0, and — + — = —1. The first 

a 2 o* a* o* 

of these is the equation of a point ellipse and the second is that of an 
imaginary ellipse. 



100. Equation of ellipse when axes are translated. — 

Transform the equation — + — = 1 by translating the coor- 
dinate axes to a new origin at a point 0'(—h, — k), using [12], 
and we have 

(x' - A) 2 



+ ^^- 2 = l. 



This is the equation of an ellipse 
having its center at the point (h, k) 
referred to the new coordinate axes, 
and having its axes parallel re- 
spectively to the rc'-axis and the 
j/-axis, as shown in Fig. 86. 



F' 



o' 



(-/».-*) 



Fig. 86. 

If the primes are dropped, this equation becomes 



->* 



+x' 



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124 ANALYTIC GEOMETRY [§101 

M fe^C + fc^-,, 

which is a second standard form of the equation of the 
ellipse, and is a convenient form for writing the equation of 
an ellipse with center at the point (A, k) and major axis 
parallel to the x-axis. 

Similarly, the equation of an ellipse with center at (A, k) 
and major axis parallel to the y-axis is of the form 

1*1 fc^SI + fe^E-i. 

Example. — Find the equation of an ellipse with semimajor axis 5, 
semiminor axis 4, center at point (3, —2), and major axis parallel to 
the o;-axis. 

Substituting in [84], {x ~ 3)> + (y + 2)> - 1. 

Y Simplifying, 

16s* + 25y* - 96* + lOOy - 156 - 0. 
The ellipse is shown in Fig. 87. 

7 EXERCISES 

1. Write the equations of the fol- 
lowing ellipses, and plot: 

(1) Center at (3, 4), a = 5, b = 3, 
and major axis parallel to x-axis. 

Y (2) Center at (-3, -7), a - 6, 
Fio. 87. & = iV% and major axis parallel to 

jfcaxis. 

2. Find the equations of the two ellipses, each having its center 
at (5, — 4), o = 6, and 6=4; one having its major axis parallel to 
the z-axis and the other having its major axis parallel to the y-axis. 

3. Find the coordinates of the foci and the equations of the directrices 
of each ellipse of exercise 1. 

4. Find the equation of the ellipse with center at (10, 2), one directrix 
the line x = 2, and eccentricity f . 

5. Find the equation of the ellipse with center at (3, 4), major axis 
parallel to x-axis, and passing through the points (—2, 4) and (3, 0). 

6. Find the equation of the ellipse having its center at (4, 2), major 
axis parallel to the y-axis, semimajor axis 6, and passing through the 
point (8, 4). 



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§101] THE ELLIPSE AND CERTAIN FORMS 125 

101. Equation of the form Ax 2 + Cy 2 + Dx + Ey + F =0. 

Every equation of the form Ax 2 + Cy 2 + Dx + Ey + F = 0, 
where A and C have like signs but different values, represents 
an ellipse with axes 'parallel to the coordinate axes. 

Proof— Given Ax 2 + Cy 2 + Dx + Ey + F = 0. 

Completing the squares in x and in y, 

D\ 2 , nl , E\ 2 CD 2 + AE*-4ACF 



A i* + U) + C (" + £) " 



4AC 
Dividing by the second member of this equation, 

(*+2z) 2 ( y+ &Y t 

CD 2 + AE 2 - 4ACF + CD 2 + AE 2 - 4ACF " lm 
AA 2 C 4AC 2 

This is in the form of [34] if A<C where 

D . tf t CD 2 + AS 2 - 4ACF 

A = ~2A' * = ~2C' a = il^C ' and 

CD 2 + AE 2 - 4ACF 
6 " IAC 2 

and therefore represents an ellipse with axes parallel to the 
coordinate axes if A and C have like signs so that 44 2 C and 
AA C 2 have like signs. It is of the form of [34J if A <C. 

From the preceding, it follows that the equation of an ellipse 
in the form Ax 2 + Cy 2 + Dx + Ey + F = can be trans- 
formed into one of the forms [32] or [33] by a suitable trans- 
lation of the coordinate axes, the new origin being at the 

point (~^-^)- 

Example 1. — Transform to the second standard form, the equation 
of the ellipse 24a* + 49y* - 96s + 294y - 639 = 0, find the coordinates 
of the center, foci, and vertices, the length of the semimajor and semi- 
minor axes, and the equations of its directrices. Plot. 
SolvJUon. — Completing the squares in x and in y, 

24(s* - 4* + 4) + 49(y* + Gy + 9) = 639 + 96 + 441, 
or 24(z - 2)* + 49(y + 3)* = 1176. 

Dividing by 1176 and putting in the form of [84], 
(s - 2)* (y 4- 3)' - 
49 T 24 



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126 



ANALYTIC GEOMETRY 



[§101 




+*.x 



This is an ellipse (Fig. 88) with center at the point C(2, —3) and 
axes parallel to the coordinate axes. The semi axes are a — 7, and 
b = 2\/S. 

The eccentricity e = ^° — — = =• 
a 7 

The distance from the center to the foci is ae = 5, and the foci are 
>(7, -3)and*"(-3, -3). 

The vertices are 7(9, -3) and 7'(-5, -3). 

The distance from the center to the directrices is - = fy and the 

e 

equations of the directrices are x — y 

and x = — V- 

The ellipse is as shown in the figure. • 

Example 2. — Find the equation of 

the ellipse whose axes are parallel to 

the coordinate axes and which passes 

through the points (—2, 7), (2, 4), 

(-2,1), and (-6,4). 

Solution. — The required equation 

is of the form 

Ax* + Cy* + Dx + Ey +F = 0. 

If this is divided by A the equation is of the form 

x s + C'y* + Jyx + E'y + F' - 0, 

and therefore contains only four arbitrary constants, which can be found 

from four equations. 

Dropping the primes and substituting the coordinates of the four 

given points, 

4 + 49C - 2D + 7E + F - 0, 

4 + 16C + 2D + 4JS + F = 0, 

4 + C-2D+E+F = 0, 

36 + 16C - 6D + \E + /? = 0. 

Solving, C-V»I>-4,J£» -H*' ^ = H 1 - 
The required equation is 

x* + W + 4* - *Vy + H* = o, 

or 9z 2 + 16y 2 + 36* - 128y + 148 = 0. 

Example 3. — Translate the coordinate axes so that the equation of the 
ellipse 4s 2 + 9y 2 - 24s - 36y + 36 = is in the form [32]. 

Solution. — Completing the squares in x and in y f 

4(s 2 - 6s + 9) + 9(t/ 2 - 4y + 4) = -36 + 36 + 36. 

Whence fe^ + <L^»! _ i. 



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§102] 



THE ELLIPSE AND CERTAIN FORMS 



127 



Putting a; — 3»s / , or x = x 7 + 3 and y — 2 -» y 7 , or y = y* + 2, 
9 



-V2 




**X 



Fig. 89. 



This is of the form [32], and is an ellipse referred to coordinate axes 
that are parallel to the old coordinate 
axes, and with the new origin at the 
point (3, 2). 

The ellipse is as shown in Fig. 89. 

The transformation could evi- 
dently be made by substituting 
x = x 1 + h and y = y f + k } and- 
proceeding as in the example of 
article 89. 

EXERCISES 

1. Express the equations of the following ellipses in the form [34] or 
[34i], find the coordinates of the centers, foci, and vertices, the lengths 
of the semimajor and semiminor axes, and the equations of the directrices. 
Plot each. 

(1) 7z* + 16y* + 14a? - 64t/ - 41 - 0. 

(2) 8x* + 4y 2 - 64^- Sy + 68 - 0. 

(3) 4z* + 9y 2 - 8s + ISy + 12 - 0. 

(4) 8s* + 9y 2 + 16* - 54t/ - 1 = 0. 

2. Transform 6s 2 + 7y 2 — 36s + 14y -f 53 = to new axes parallel 
respectively to the old axes, with the new origin at (3, —1). 

3. Transform each of the ellipses of exercise 1 to the form [32] or 
[33],. find the coordinates of the foci, and the equations of the directrices 
referred to the new coordinate axes. 

4. Find the equation of the ellipse with major axis parallel to the 
a5-axis, and center at the point (—3, 4), eccentricity $» and passing 
through the point (6, 9). 

5. Find the equation of the ellipse with one focus at the point (6, 2), 
corresponding directrix the line x = 12, and eccentricity i- 

6. Transform the following equation to one in which there are no 
x and y terms, and plot: 9s 2 + 12y 2 — 18z — 72y + 9 = 0. 

7. Find the equation of the ellipse with eccentricity i, a focus at 
the point (2, 0), and the corresponding directrix the line x + 2 ■» 0. 

102. Equation of ellipse when axes are rotated. — In article 
90 it was seen that when the coordinate axes were rotated 



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128 



ANALYTIC GEOMETRY 



[§102 



through an angle <p, a term in xy appeared in the equation of 
the parabola. Likewise if the equation of an ellipse with axes 
parallel to the coordinate axes, Ax 2 + Cy 2 + Dx + Ey + F = 0, 
is transformed by using formulas [13], the equation takes the 
form Ax* + Bxy + Cy 2 + Dx + Ey + F = 0. This is 
the most general form of an equation of the second degree in 
x and y, where B* - 4AC<0. (See Art 122.) 

Conversely, starting with an equation containing an xy- 

term, rotation through a properly 
chosen angle will cause the xy-term 
to disappear by having its coeffi- 
cient zero. 

Example 1. — Transform the equation 
9x* + 16y 2 - 36a; - 96y + 36 = 0, by ro- 
tating the coordinate axes through an 
angle of 30°. Sketch the ellipse. 
Solution. — Using formulas [13], 
x = x' cos 30° - if sin 30° - 
ly/S* - W - i( a/3*' - V'), 
and y = x> sin 30° + \f cos 30° = \x' + \y/%\f - W + y/ltf). 

Substituting these values in the given equation and simplifying, 
43« , *+14v^a;yH-57y , *-24(3VS+8)x , -24(8V3--3)y , -|- 144 = 0. 
The ellipse and the two sets of coordinate axes are sketched in Fig. 90 

EXERCISES 

1. Transform the following equations by rotating the coordinate axes 
through the angle given in each case: 




f+-X 



Fig. 90. 



wS+B-i- 



<P = 45° 



(2) 16s* + 9y* - 144. <p - 60°. 

(3) 3ftr* + 4y» = 144. <p = 90°. 

(4) 2x* + 3y* - 4* + Zy - 10 = 0. *> - 30°. 

(5) s* + xy + y 2 + 2x + y + 2 = 0. *> = 45°. 

(6) 6s* + 4sy + 6y* + 5s - Sy = 0. *> = tan" 1 J. 

2. Transform the following equation to the standard form by rotating 
the axes: 29s 2 + 16sy + 41y* - 45 = 0. Sketch the ellipse with both 
sets of axes. 

3. Simplify the following equation by first translating the axes to 
remove the z-term and the y-term, then by rotating through an angle 



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§103] 



THE ELLIPSE AND CERTAIN FORMS 



129 



that will remove the xy-term. Sketch the curve and the three sets of 
coordinate axes: x* + xy + y* + 2x + 3y — 3 =0. 

103. Equation of ellipse in polar coordinates. — In a manner 
similar to that of article 91, the equation of the ellipse in 
polar coordinates may be derived. 

EXERCISES 

1. Derive the equation of an ellipse with the pole at the focus to 
the right of its corresponding directrix, and the polar axis perpendicular 
to the directrix. Also derive the equation when the focus is taken at 
the left of its corresponding directrix. Let p equal the distance from the 
focus to the directrix. 

2. Transform the results of exercise 1 to rectangular coordinates, 
and change to the standard form by translation of axes. 

3. Derive the polar equation of an ellipse, the pole being at a focus, 

Cx — c\ ' i/' 
by starting with the equation 5 \- p- = 1, and then putting 

x = p cos 0, y = p sin 0, c — ae, and b* — (1 — e 2 )o 2 ; finally solving the 
quadratic equation for p. 

4. Derive the polar equation of an ellipse, the pole being at the center 
and the polar axis along the major axis. 

5. Show that pe in exercise 1 is one-half the latus rectum* 

104. Construction of an 
ellipse. — First method. — The 
length of the major axis 2a and 
the foci F and F f are supposed 
to be known. 

On a drawing board fasten 

the ends of a string of length 2a 

at F and F', Fig. 91. Place a 

pencil point, P, in the string 

and move it about keeping the 

string taut. Then the point P will generate an ellipse. 
This construction depends upon the following: 
Theorem. — The sum of the distances from any point on an 

ellipse to its foci is constant and equal to the major axis. 

< This may be proved as follows: In Fig. 92, from the 

definition of an ellipse, 




Fig. 91. 



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130 



ANALYTIC GEOMETRY 



[§104 



and 



PF' - e-N'P = e(^ + x) = a + ex, 
PF = e-PN = eg-z) = a- 



err. 



Adding, PF' + PF = 2a = major axis. 

Second method. — The major axis, 2a, and the minor axis, 
26, are supposed to be known, as well as the position of the 
center and direction of axes. 

With the center of the ellipse as a center describe two 
circles of radii a and 6 respectively, Fig. 93. Draw any radius 
intersecting the inner circle in R and the outer circle in Q. 
Through R draw a line parallel to the major axis, and through 
Q a line parallel to the minor axis. Then the point P where 
these lines intersect is a point on the ellipse. In this manner 
any number of points on the ellipse can be determined. 

That the point P is on the 
ellipse with its major axis on 
the x-axis can be proved as 
follows: 



N 



D i 


J 


\f' o 


F J 

€ a 


D' - ~ 


< - > 



•*x 



Fig. 92. 




Fig. 93. 



y z 



Equation of ellipse is — 2 + ~ = 1. 
a o 



If is the angle XOQ, the coordinates of P are 

x = OM = a cos B y 

and y = MP = b sin 6, 

Substituting in the equation of the ellipse, 

a 2 cos 2 6 , 6 2 sin 2 6 n , . 

— ~2 h r^ — = cos 2 6 + sin 2 = 1. 



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§105] THE ELLIPSE AND CERTAIN FORMS 131 

Hence the equation is satisfied and the point P is on the 
ellipse. 

EXERCISES 

1. By the second method construct an ellipse having semiminor and 
semimajor axes 1 in. and 1J in . respectively. 

2. Prove that the projection of a circle upon a plane making an acute 
angle with the plane of the circle is an ellipse. 

APPLICATIONS 

105. Uses of the ellipse. — The ellipse is involved in many 
practical considerations, as well as being frequently used in 
mathematics and its applications. 

It was believed by tfie ancient Greeks that the sun was the 
center of the universe in which we live. Kepler (1571-1630) 
stated that the orbits of the planets are ellipses. Newton 
(1642-1727) showed that the law of gravitation determines 
the orbits to be ellipses. 

In architecture, because of the beauty of its form, the 
elliptic arch is frequently used. Some noted structures were 
built in the form of an ellipse. The Colosseum at Rome was 
of this form. 

In bridge structures, many of the most noted stone-arch 
bridges of the world are elliptical. 

In machinery, elliptical gears are often used where change- 
able rates of motion are desired, as in shapers, planers, and 
slotters where the cutting speed is less than the return motion. 

In the study of electricity and mechanics, the ellipse is 
frequently used. 

EXERCISES 

1. The Colosseum at Rome is in the form of an ellipse 615 ft. long 
and 510 ft. wide. Find the equation of the ellipse and the position 
of the foci. 

2. A stone-arch of a bridge has a span of 200 ft. and a height of 42 ft. 
The arch is in the form of a semi-ellipse. Find the equation of the 
ellipse and the position of the foci. 



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132 ANALYTIC GEOMETRY [§105 

3. In exercise 2, find the heights of points 50 ft. and 25 ft. from one end 
of the arch. 

4. In considering equipotential surfaces in electricity, the equation 

. , x + — ; x = 1 is used. If a > b and X denotes an arbitrary con- 
a* + A o* + A 

slant, such that X > — b* show that the equation represents a system 

of ellipses haying the same foci. 

5. An arch is in the form of a semi-ellipse with major axis horizontal. 
The span is 80 ft. and the height is 30 ft. Find the distance of the 
arch below the level of its top for each 10 ft. of the span. 

6. The earth's orbit is an ellipse with the sun at one focus. The 
major axis is 185.8 million miles and the eccentricity is about A* 
Find the difference between the greatest and the least distance from the 
earth to the sun. 

7. Show that, if two equal elliptical gears turn on mountings at 
corresponding foci, they are always in contact. 

8. If two equal elliptical gears have major axes and minor axes of 
12 in. and 8 in. respectively, and revolve once in 10 seconds, find the 
greatest and the least linear speed of a point on the driving ellipse. 

Suggestion. — Use the greatest and the least radius on which a point 
is turning. 

The driving gear has uniform angular velocity, and the mountings 
are at corresponding foci. 

'• GENERAL EXERCISES 

1. Find the equation of an ellipse in the form of [32] having the sum 
of its axes 20, and the difference 4. 

2. Find the equation of an ellipse in the form of [33] if its major axis 
is 24, and its minor axis is equal to the distance between the foci. 

3. Find the equation of an ellipse in the form of [32] if the minor 
axis is 12, and the distance between the foci is 12. 

4. Find the equation of the ellipse in the form of [33] in which o = 8, 
and the foci bisect the semi major axes. 

Find the semi-axes, coordinates of foci, eccentricity, and the equa- 
tions of the directrices of each of the following ellipses: 

5. 16x 8 + 9y 8 = 144. 

6. 24x 8 + 36y 8 = 864. 

7. 16s 8 + 25y 2 - 64a; + lOOy - 236. 

Transform each of the following equations to axes parallel respec- 
tively to the old, the new origin being at the point given in each case. 
Plot the curve and both sets of axes. 



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§105] THE ELLIPSE AND CERTAIN FORMS 133 

8. 9x* + 4y* + 3ftr - 24y + 36 - 0. (-2, 3). 

9. 25s 8 + 16V + 50x + 32y - 359 - 0. (-1, -1). 

10. Derive an equation that will represent all ellipses having foci at 
the points (3, 0) and (-3, 0). 

11. Derive the equation of the ellipse with a focus at (3, 1), eccentricity 
equal to }, and with Zx — 4y + 6 «■ as directrix. 

12. Show that the latus rectum of an ellipse is a third proportional to 

x s y 1 
the two axes. Find the latus rectum of the ellipse — + — : — 1 by 

this method. 



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CHAPTER VIII 

THE HYPERBOLA AND CERTAIN FORMS OF THE 
SECOND DEGREE EQUATION 

106. The equation of the hyperbola. — By the definition of 
article 81, the hyperbola is the locus of a point whose distance 
from a fixed point, the focus, is to its distance from a fixed 
straight line, the directrix, in a constant ratio e, greater than 1. 

The method used in deriving the equation is exactly the 
same as that for the ellipse, Art 95. In Fig. 94, let F be the 
focus and D'D the directrix. Choose as x-axis the line X'X 
through F and perpendicular to D'D at R. 




Since e > 1 there are two points V and V on X'X such 

VF V'F 

that ^y = e and y^ = e. Hence the points V and V are 

on the locus. 

Choose 0, the point midway between V and V' y as origin, 
and Y'Y f parallel to D'D, as t/-axis. 

Let the length of V'V = 2a. Then V'O = OV = a. 

134 



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§106] THE HYPERBOLA AND CERTAIN FORMS 135 

As with the ellipse it is necessary to find the equation of 
the directrix and the coordinates of the focus. 

From the definition of the hyperbola, 

VF = e RV, or OF - a = e(a - OR), (1) 

and VF = eV'R, or OF + a = e(a + OR). (2) 

Subtracting equation (1) from equation (2), 

2a = 2eOR, or OR = -• 

Then the equation of the directrix is x = — 

Adding equations (1) and (2), 

20F = 2oe, or OF = ae. 

Then the coordinates of F are (ae, 0). 

To derive the equation of the hyperbola, let P(x, y) be any 
point on the locus, join F and P, and* draw iVP perpendicular 
to D'D. 

By definition, FP = e i\TP. 

But FP ** V(* - <*0 2 + 2/ 2 , and NP = x - j. 

Then V(s - a*) 2 + y 2 = e(s T fj • 

Squaring and arranging, this equation becomes 
x 2 y 2 

a 2 ~ a 2 (e 2 - 1) = 
Since e>l, a 2 (e 2 — 1) is positive. Let it be represented by 
b 2 and the equation of the hyperbola is 

x 2 v 2 

This is a standard form of the equation of the hyperbola, 
and is the form in which the hyperbola is usually written. 
Its simple form is due to the choice of the coordinate axes. 
A different choice of axes would give a less simple form of the 
equation, but the locus would be unaltered. 

Since b 2 = a 2 (e 2 -!),« = ^ a% + b *- 



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136 



ANALYTIC GEOMETRY 



[5107 



Equation [35] is the required equation for it has been proved 
true for every point on the hyperbola, and it can be readily 
proved that it is not true for any point that is not on the locus. 

107. Shape of the hyperbola. — The shape of the hyperbola 
and its position relative to the coordinate axes can be readily 

x 2 t/ 2 
determined from the equation — 2 — 4- t = 1. 

Solving for x, 

Solving for y, 

(1) For all values of y, x has two real values, numerically 
equal but opposite in sign. For all values of x such that 



x - ± j^Vbf+V*. 



y hs ± -V x* — a 2 




'(-a.O) o 




(a.0)l (<w.O) 



+x' 



Fig. 95. 

x 2 — a 2 >0, y has two real values, numerically equal but 
opposite in sign. When x 2 = a 2 , y = 0. Hence the curve is 
symmetrical with respect to both coordinate axes and the 
origin, and its intercepts on the x-axis are a and —a. 

(2) For all values of x such that x % — a 2 <0, y is imaginary, 
but no value of y will make x imaginary. 

(3) As x increases from + a or decreases from — a, the positive 
values of y increase and the negative values of y decrease. 

The hyperbola has the shape shown in Fig 95. 



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§108] 



THE HYPERBOLA AND CERTAIN FORMS 



137 



108. Definitions. — The center of symmetry of the hyperbola 
is called the center of the hyperbola. 

The line through the focus and perpendicular to the directrix 
is called the principal axis of the hyperbola. 

The points in which the hyperbola intersects the principal 
axis are called the vertices of the hyperbola. 

The portion of the principal axis lying between the 
vertices is called the transverse axis of the hyperbola. Its 
length is 2a. 

The conjugate axis of the hyperbola has a length 26, is 
perpendicular to the principal axis, is bisected by it, and 
passes through the center. 

The chord of the hyperbola through the focus and perpen- 
dicular to the principal axis is called the latus rectum. Its 

26* 
length is — > for the abscissa of the focus is ae, and when 

r 

E 



a 



x = ae,y 

109. Second focus and 
second directrix. — The hy- 
perbola -j — p = 1 has a 

second focus at the point 
(— ae, 0), and a second direc- 
trix which is the line x = 



a 
e 




The proof is similar to that 

of article 98 for the ellipse and is left as an exercise for the 

student. 

In Fig. 96, F and F 9 are the foci, and the lines D'D and E'E 
are the directrices. 

110. Hyperbola with transverse axis on the y-axis. — The 
equation of an hyperbola whose transverse axis is on the 
y-axis and whose center is at the origin is obtained by inter- 



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138 



ANALYTIC GEOMETRY 



[5110 



changing x and y in the work of article 106. 
then 

y 2 x 2 



The equation is 



[36] 



J -„ - ,-. = 1. 




Here the transverse axis is 2a; the conjugate axis is 26; the 

coordinates of the vertices are 
(0, ±a); the coordinates of the 
foci are (0, ±ae); and the equa- 
tions of the directrices are y = +-• 

6 

(See Fig. 97.) 

EXERCISES 

1. In each of the following hyperbolas, 
find the length of the transverse axis 
and the conjugate axis, the coordinates 
of the foci, and the equations of the 
directrices. Sketch each hyperbola. 

(6) 9y* - 25s* - 225. 

2. Write the equation of the hyperbola with center at the origin, and 
transverse axis on the s-axis, having given: 

(1) p * - 6, b - 4. (4) b = 3, oe = 5. 

(2) a = 4, e »» 2. (5) a - 9, e - $• 

(3) 6 = 8, ae - -%S/5- (6) 6 - 6, at - \/85. 

x* y* 

3. In the hyperbola - — — = 1, find the value of y when x » 3, 

when x = 5, when 3 = 2. 

4. Find the length of the latus rectum of the hyperbola — — — = 1. 

36 lo 

y 2 x 2 
Of the hyperbola — - — — = 1. 
a 2 b 2 

5. Find the equation of an hyperbola with transverse axis on the 
x-axis, center at origin, and passing through the points (6, 4) and ( —3, 1). 

6. Find the equation of the locus of a point moving so that the differ- 
ence of its distances from the points ( ±6, 0) is 8. 



Fig. 97. 

<■>§-?«->■ 

(2) x - - J?- . 1 
W 36 100 *' 

(3) 16a; 1 - 9y l = 144. 



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§111] 



THE HYPERBOLA AND CERTAIN FORMS 



139 



7. Derive equation [36] from [35] by rotating the coordinate axes 
through an angle <p = 90°. 

8. Find the semi-axes, eccentricity, and the latus rectum of each of 
the following hyperbolas: 



(1) ii- 






i. 



(4) - " V* - m. 



(2) 4x* - 3y* - 24. (5) px* - qy* « pq. 

(3) 16x* - y* - 16. (6) s a - qy* - «. 

9. Find the semi-axes, coordinates of foci, eccentricity, and equations 
of directrices of each of the following hyperbolas: 

(1) 16s 8 - 9y* - 144. (2) 24x 8 - 36y 8 - 864. 

10. Find the equation of an hyperbola with transverse axis on the 
y-Sixia, center at the origin, eccentricity equal to 2, and passing through 
the point (3, 2). 

x 8 y* 

11. Assume the equation --- = 1 of the hyperbola, and show 

a 2 o 2 

that the difference of the distances of any point on it from the foci is 2a. 

12. Show that the latus rectum of an hyperbola is a third proportional 
to the two axes. 

13. What does the equation x % — y % = 16 become when the coordinate 
axes are rotated through an angle <p = —45°? 

14. Find the equation of an hyperbola if its center is at the origin, 
transverse axis is 24, and the distance between its foci is 32. 

15. Find the equation of an hyper- 
bola if its center is at the origin, 
transverse axis is 24, and its con- 
jugate axis equals one-half the dis- 
tance between its foci. 

111. Asymptotes. — In Fig. 98, 
P'P is a line passing through the* 
center of the hyperbola and in- 
tersecting the curve in P' and P. 
If P is made to move off to 
infinity along the curve, the line 
P'P \ continually passing through v A * 
the center, will turn about and 
will approach one of the two 
lines A 1 A or B'B. These lines are called the asymptotes 1 of 
x the hyperbola. 

1 This is not the general definition for asymptotes, but is true for the 
hyperbola. 




+x 



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140 



ANALYTIC GEOMETRY 



[§111 



The equation of any line P'P through the origin is y =* mx. 
The coordinates of its intersection with the hyperbola 



6* 



1 are found by solving the equations as simultaneous 



equations. 
Solving for x, 



x = ± 



ab 



Vb* - a*m 2 ' 



Now as P moves off to infinity along the curve x becomes 
infinite. Therefore the denominator of the fraction must 
approach 0. 

This gives 6 2 — a*m 2 = 0, or m = ± — 



Hence the equations of the asymptotes are 
[37] 



y - -x, and y -x. 



These equations can be 
combined into the single 
equation 

a* b* U * 
The conjugate axis B'B, 
Fig. 99, can now be brought 
into a closer relation to the 
hyperbola. If through the 
extremities of B'B lines are 
drawn parallel to the transverse axis, and through the ex- 
tremities of the transverse axis V'V lines are drawn parallel 
to the conjugate axis, a rectangle is formed with its diagonals 
on the asymptotes of the hyperbola. 

It can readily be shown that if the transverse axis of the 
hyperbola is on the #-axis, the equations of the asymptotes are 




y 



and 



V--5*. 



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§112] 



THE HYPERBOLA AND CERTAIN FORMS 



141 



By the help of the asymptotes, a simple and fairly accurate 
method for sketching an hyperbola is as follows: Locate the 
vertices and draw the asymptotes, then draw the hyperbola 
so that the curve continually 
approaches the asymptotes as it 
nfoves off toward infinity. 

Example. — Sketch the hyperbola 
16x» - 25y* - 400. 
First put 16s* - 25y* - 400 in the * 

Then a = 5, b = 4, the foci are at 
the points (5, 0) and (—5, 0), and the 
equations of the asymptotes are 
y = ix and y - -£r. p^. 100# 

The curve is as shown in Fig. 100. 

112. Conjugate hyperbolas. — Two hyperbolas that are so 
related that the transverse axis of each is the conjugate axis 




Fig. 101. 



of the other, both in magnitude and in position, are called 
conjugate hyperbolas. 

Thus, ^= — r; = 1 and r? — ^= =* 1, Fig. 101, are conjugate hyperbolas. 



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142 ANALYTIC GEOMETRY [§113 

From article* 111, it is seen that the asymptotes of each are 

y = ±-z. Therefore two conjugate hyperbolas have the same 

asymptotes. 

The formula 6 2 = a 2 (e 2 — 1) can now be readily interpreted 
geometrically. For in right triangle OVN, Fig. 101, OV = a, 
VN = OB = 6, and ON = OF = ae. 

113. Equilateral hyperbola. — If a = 6, the hyperbola 

x 2 y 2 

— 2 ;— t- " 1 becomes x 2 — j/ 2 = a 2 . This is called an 

equilateral hyperbola. 

The equations of its asymptotes are y = ±x, and are 
evidently perpendicular to each other and make angles of 45° 
with the axes of the hyperbola. 

An equilateral hyperbola is also called a rectangular 
hyperbola. 

It may be noted that the equilateral hyperbola is the 
simplest of hyperbolas, just as the circle is the simplest of 
ellipses, being the ellipse in which the major axis and minor 
axis are equal. 

EXERCISES 

1. Find the equations of the asymptotes and sketch the curve for each 
of the following hyperbolas: 

(1) * _ £ ' . 1 ( 4 ) Ml » £* = 1 
K } 18 12 l ' w 25 16 * 

(2) 9z* - 18y* = 16. (5) x* - y* - 12. 

(3) 9s 2 - 1%* = -16. (6) x* - y 2 - -12. 

2. One of two conjugate hyperbolas is 12z 2 — I62/ 2 = 192, find the 
other. Find the coordinates of the foci and the equations of the direc- 
trices of each. 

3. Show that the four foci of two conjugate hyperbolas, and the four 
points of intersection of the tangents at their vertices, all lie on a circle 
whose center is at the common center of the two hyperbolas. 

4. Show that the eccentricity of an equilateral hyperbola is \/2. 

6. Transform the equation of the equilateral hyperbola x* — y % — a*, 
by rotating the coordinate axes through an angle <p = —45°. This 
refers the hyperbola to its asymptotes as axes. 



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§114] THE HYPERBOLA AND CERTAIN FORMS 143 

6. Find the equation of the hyperbola whose vertices are at ( ±4, 0) 
and the angle between whose asymptotes is 60°. 

7. If ei and e a respectively. are the eccentricities of two conjugate 

hyperbolas, show that ae x « bet and that — = H — = -*» 1. 

ei z e** 

8. Plot the equilateral hyperbolas x* — y % = a* and y* — x* = a* 
and locate their foci. With the same coordinate axes plot the circle 
x % + y 2 = 2a*. Also plot x 1 — y* = on the same set of axes. 

9. Prove that in any hyperbola the distance from a focus to an asym- 
ptote equals the semi-conjugate axis. 

10. Prove that in any hyperbola the distance from the center to the 
foot of the perpendicular from a focus to an asymptote equals the semi- 
transverse axis. 

11. Find the value of b in order that the line y = 2x -f b shall be 

• x* y* 

tangent to the hyperbola = 1. 

9 4 

12. Find the value of m in order that the line y = mx -f 2 shall be 

x 1 y* 
tangent to the hyperbola — — — = 1. 
16 9 

114. Equation of hyperbola when axes are translated. — 

By a method identical to that of article 100 for the ellipse, 
the equation 

[38] — - 2 ^ 1 

is found for the hyperbola with its center at the point (h, fc), 

and whose transverse axis is parallel to the x-axis. This is 

a second standard form of the equation of the hyperbola. 

If the transverse axis is parallel to the t/-axis the equation is 

raai (y - k) 2 (* - h) 2 _ ± 
[38J — - 2 p i- 

EXERCISES 

1. Write the equations of the following hyperbolas: ' 

(1) Center at (4, — 3) , a =5,6= 3, and transverse axis parallel to oj-axis. 

(2) Center at (—6,-2), a = 2, b =4, and transverse axis parallel 
to y-axis. 

2. Find the coordinates of the vertices and the foci, and the equations 
of the directrices of each hyperbola of exercise 1. 

3. Find the equation of the hyperbola with center at (—2, 7), one 
directrix the line y = 5, and eccentricity equal to £ . 



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144 ANALYTIC GEOMETRY [§115 

4. Find the equations of the hyperbolas that are conjugate hyperbolas 
with those of exercise 1. 

5. .Find the equations of the asymptotes of the hyperbolas of exercise 1 . 

115. Equation of the form Ax 2 + Cy 2 + Dx + Ey + F - 0. 
Every equation of the form Ax 2 + Cy 2 + Dx + Ey + F = 0, 
where A and C have unlike signs, represents an hyperbola with 
axes parallel to the codrdinate axes. 

Proof. — In a manner identical to that of article 101 the 
equation takes the form 



~T S1T\9 I A ET9 A A riTZ ™" *• 



CD 2 + AE 2 - 4ACF ' CD 2 + AE 2 - 4ACF 
4tA 2 C 4AC 2 

This is of the form of [38] or [38i] for the denominators 
have unlike signs since A and C are unlike in sign and therefore 
4tA 2 C and 4tAC 2 are unlike in sign. 

If the second denominator is negative, the transverse axis 
is parallel to the X-axis. If the first denominator is negative, 
the transverse axis is parallel to the y-axis. 

From the preceding proof it follows that the equation of an 
hyperbola in the form Ax 2 + Cy 2 + Dx + Ey + F = can 
be transformed into the standard forms, [35] or [36], by a 
suitable translation of the coordinate axes, the new origin 

being at the point (-^Z' ~"2C/ 

Example.— Express the hyperbola 3&r* - 25y* + 21ftc + 100y-676 =0 
in the form of [38]. What are the coordinates of its center, foci, and 
vertices; the lengths of the semi-axes; and the equations of its directrices 
and asymptotes? Plot. Finally, translate the codrdinate axes so as to 
change to the form [35] and answer the same questions with reference to 
the new axes. 
Solution. — Completing the squares in x and in y, 

36(3* + Ox + 9) - 25(y* - 4y + 4) - 676 + 324 - 100, 
or 36(s + 3) 8 - 25(y - 2)* - 900. 

Dividing by 900 and putting in the form [38], 
. (s+3)* (y-2)» _\ 
25 36 



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Ins] 



THE HYPERBOLA AND CERTAIN FORMS 



145 



This is an hyperbola with its center at the point C(— 3, 2) and trans- 
verse axis parallel to the z-axis. 
The semi-axes are 5 an d 6. 

a 5 

The distance from the center to the foci is ae =•- VST, and the foci are 
F(-3 +V6T 2) and *"(-3 - V$l, 2). 



The eccentricity e 




Fiq. 102. 



The vertices are 7(2, 2) and V'(-8, 2). 

The distance from the center to the directrices is 



VQ1 



and the 



5 5 

equations of the directrices are x = — 3 H 7=. and x ■■ — 3 ;=• 

V81 V61 

The asymptotes have slopes of $ and — # respectively, and pass through 
C(— 3, 2). Their equations are by [15], 

y - 2 - !(* +3). and y - 2 = -£(x + 3), 
or 6x - 5y + 28 - 0, and 6a; + by + 8 = 0. 

The hyperbola is as shown in Fig. 102. 

To change to the form of [36], put x + 3 = x 1 and y — 2 — yf. Then 

x'* y'* 
1 becomes — — — = 1, referred 
25 36 ' 



the equation 



(x + 3)» (y - 2)» 



25 36 

to CX< and. CY' as axes. The center is C(0, 0); foci are F(V6% 0) 
and F'(-\/6% 0); vertices are F(5, 0) and V(-5, 0); equations of 



directrices are x ■» ■ 



V6T 

6* — 5y — and 6x + by 
10 



and a; — — 



\/61 



; and asymptotes are 



0. 



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146 ANALYTIC GEOMETRY [§116 

EXERCISES 

1. Express the equations of the following hyperbolas in the form 
of [88] or [88i]. Find the coordinates of the centers, foci, and vertices; 
the lengths of the semi-axes; and the equations of the directrices and 
asymptotes. Sketch each curve. 

(1) 9s* - 16y* - 108x + 96y + 36 - 0. 

(2) 16y* - x* - 6x - SOy + 75 - 0. 

(3) 8x* - 28y* - Sx - 2Sy - 61 - 0. 

(4) Sx* - 9y* - 16s + 64y - 1 - 0. 

(5) 3y* - 4s* - 16x - 24y - 52 *- 0. 

2. Transform 9s* - 25y* + 54s + lOOy + 206 = by translating to 
new coordinate axes parallel respectively to the old axes, with new 
origin at (—3, 2), and sketch the curve. 

8. Transform each of the hyperbolas of exercise 1 to the form of 
[86] or [36]. Find the coordinates of the foci, and the equations of the 
directrices referred to the new coordinate axes. 

4. Find the equation of the hyperbola with conjugate axis parallel 
to the x-axis, center at the point (—3, 4), eccentricity f> and passing 
through the point (9, 4 + 8\/3). 

6. Find the equation of the hyperbola whose axes are parallel to 
the coordinate axes and which passes through the points (3, 4), (—7, 4), 
(8, 4 + 4\/3), and (-12, 4 - 4\/3). 

6. Find the equation of the hyperbola having a focus at (6, 2), a 
directrix the line s — 12 = 0, and e = 2. 

116. Equation of hyperbola when axes are rotated. — In 

like manner to that for the parabola (Art. 90) and the ellipse 
(Art. 102), the equation Ax 2 + Cy 2 + Dx + Ey + F = 0, 
which is that of an hyperbola with axes parallel to the coordi- 
nate axes, is transformed by using equations [13] to the form 
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. This is the most 
general form of the second degree equation in x and y r where 
B 2 -4AC>0. (See Art. 121.) 

Conversely f starting with an equation containing an xy-terin, 
rotation through a properly chosen angle will cause the 
xy-term to disappear by having its coefficient zero. 

EXERCISES 

1. Transform the following equations by rotating the codrdinate axes 
through the angle given in each case: 



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§1171 THE HYPERBOLA AND CERTAIN FORMS 147 

(l)s*-y» = 16. *--45°. (2) ~ - jj = 1. * - 90°. 

(3) sy - 8. *> = 45°. (4) 9y» - 16a;* - 144. <p - 60°. 

(5) x* - 4sy + y* + 9 = 0. <p - 45°. 

(6) a; 2 - 4sy + y* + 4\/2* - 2V% + 11-0. *> = tan"* 1. 

. 2. Transform the following equations into the standard form rotating 
the axes through a proper angle, and sketch the curve in each case: 

(1) x* + 4xy + y* - 16. 

(2) 9x* + U\/Sxy - 5y» - 48 - 0. 

3. Simplify the following equation by first translating the axes 
to remove the x-term and the y-term, then by rotating through an 
angle that will remove the xy-term. Sketch the curve and the three 
sets of coordinate axes. 

x* + 2xy - y* + Sx + 4y - 8 - 0. 
Suggestion.— Find <p = 22J°. Then use sin 22J° = i y/2 - y/2 and 
cos 22i° - i V2 + V2. 

4. In the hyperbola of exercise 3 find the coordinates of the center 
and the foci, and the equations of the transverse and conjugate axes, 
and asymptotes, referred to the original axes. 

117. Equation of hyperbola in polar coordinates. — Here 
the procedure is similar to that for the parabola and ellipse, 
and the equations of these three conies should be compared. 

EXERCISES 

1. Derive the equation in polar coordinates of the hyperbola with 
the pole at the left hand focus and polar axis along the transverse axis. 
Let p equal the distance from the focus to the directrix. 

2 V Plot the following hyperbolas and draw the asymptotes: 

(1) p = 1 - e sin (2) p "" 1 + e sin 

3. Transform x* + y* = e*(x + p} 2 into polar coordinates. 

4. Show that in the equation of the hyperbola, p = ^ - -> 

^ ' 1 — e cos 

the inclination of the asymptotes is cos"" 1 ( ±-\ • 

6. Find the polar intercepts of the conic p = ^ -, and show 

* K 1 — e cos 6' 

that the transverse axis of the hyperbola is t y v and the major axis of 
the ellipse is 1 __ 2 » 

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X48 



ANALYTIC GEOMETRY 



[§118 



6. Transform the polar equation p'cos20 — a 2 into rectangular coordi- 
nates, having the origin at the pole and the x-axis along the polar axis. 



7. Show that p 2 
-6* 



& 2 



is an ellipse if e < 1; and that 



■>*« 



1 — e 2 cos 2 $ 
is an hyperbola if e > 1. Sketch each curve. 



1 — e 2 cos 2 $ 

118. Construction of an hyperbola. — First method. — The 
length of the transverse axis, 2a, and the foci F and F' are 
supposed known. On a drawing board place two tacks at 
F and F\ respectively, Fig. 103. Tie a pencil firmly at point 
P near the middle of a string. Pass one part of the string 




V l 


> 


; i 


: 


x'A" 










o 


. a . 




/ I 





Fio. 103. 



Fio. 104. 



under the tack at F and over the tack at F # , and the other 
part over the tack at F'. Adjust the string so that 
PF' - PF = 2a. Hold the parts of the string firmly together 
at Q and pull downward. The point P will generate an arc 
of an hyperbola. By arranging the string properly other 
arcs of the hyperbola may be generated. 

This construction depends upon the following. 

Theorem. — The difference of the distances from any point on 
an hyperbola to its two foci is constant and equal to the transverse 
axis. 

This may be proved as follows: In Fig. 104, from the 
definition of an hyperbola, 

PF' = e : N'P = e(* + f)= ex + a, 
and PF = e-NP = e(x - -) = ex - a. 



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§118] 



THE HYPERBOLA AND CERTAIN FORMS 



149 



Subtracting, PF' — PF — 2a = transverse axis. 

Second method. — A focus, the corresponding directrix, and 
the value of e are supposed known. 

In Fig. 105, let F be the focus, D'D the directrix, X'X the 
axis through the focus and intersecting D'D in A, and let the 
lines QR and TS be drawn through A with inclinations respec- 
tively equal to tan -1 ( + e). Also draw a series of lines 
parallel to D'D. 




Fig. 105. 

Then the points P and P' of the curve, on any one of these 
parallels, are found by striking arcs with the focus as center 
intersecting the parallel lines and using as a radius the length 
M N of that particular parallel. Show why this is so. 

EXERCISES 

1. Locate a directrix and a focus and construct an hyperbola with 
e - $. With e - |. With e - \/3. 

2. Construct a parabola by the same method. 

3. Using the same method, construct an ellipse with (1) e «• }, 

(2) e - J V3, (3) e - f 



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150 ANALYTIC GEOMETRY [§119 

4. The difference of the distances of a point on an hyperbola from 
the foci is 4; and the foci are at the points (3, 0) and (—3, 0). Use 
the theorem of Art. 118 and derive the equation of the hyperbola. 

APPLICATIONS 

119. Uses of the hyperbola. — Whenever the law connecting 
two variables is an inverse variation it gives rise to the equa- 
tion xy = fc, where x and y are the variables and k is a constant. 
This relation often occurs in physics, chemistry, and engi- 
neering. 

Boyle's Law which states that for a perfect gas the pressure 
varies inversely as the volume, gives rise to the equation 

pv = k. 
This is not used so much in practical work as is some slight 
variation of it. (See Art. 124.) 

Then again, if the law governing the location of a point is 
v such as to fulfill the conditions of 

the theorem of article 118 the locus 
is an hyperbola. 

These and other applications are 
best illustrated by examples. 

Example 1. — Given 20 c.c. of air at 1 
atmosphere pressure. If the volume t; 
varies inversely as the pressure p, derive 
the equation showing the relation be- 
ptween the volume and the pressure. 
Plot the curve for values of p from 1 

" ' atmosphere to 20 atmospheres. 

Fia. 106. Solution. — Since v varies inversely as 

the pressure, pv — fc. 
When p = 1, v = 20, hence 1-20 = fc, or k = 20. 
Therefore the equation showing the relation between p and t; is 

pv - 20. 
This is the equation of an equilateral hyperbola referred to its asymp- 
totes as axes, and can be plotted, as accurately as desired, by points. 
It is plotted in the first quadrant only because both volume and pressure 
must be positive. (See Fig. 106.) 



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§119] 



THE HYPERBOLA AND CERTAIN FORMS 



151 



V 


1 


2 


4 


6 


8 


10 


15 


20 


V 


20 


10 


5 


3.3 


2.5 


2 


1.3 


1 



Example 2. — Instruments for recording sound are placed at two 
points A and B 500 ft. apart, Fig. 107. The report of a cannon is 
recorded 0.25 second earlier at A than at £. 
Find the equation of the locus of the position 
of the cannon, and plot. (Sound travels 
1120 ft. per second.) 

Solution. — Since sound travels 1120 ft. per 
second, the cannon is 0.25 X 1120 ft. = 280 ft. 
nearer A than B. 

Choose as origin the point midway between 
A and B, with x-axis through these points. 
Let P(x, y) be any position of the cannon. 

The coordinates of A and B are respectively 
(250, 0) and (-250, 0). 

Then BP - AP - 280. Fio. 107. 

OrV(-250 -aO* + y* - V(250 - *)» + y» - 280. 




Simplifying, 



= 1. 



19,600 42,900 

This is an hyperbola with a = 140, and b = 207.1. The conditions 
require, however, that the locus of the position of the cannon shall be 
the branch of the hyperbola nearer to A. 

Note. — The above example illustrates the principle made use of in 
the most accurate instruments used by the Allies in the Great War 
for locating hidden guns. Near the close of the war they were locating 
guns 10 miles distant within a radius of 5tf ft. 

EXERCISES 

1. Given 10 c.c. of air at atmospheric pressure. If the volume v 
varies inversely as the pressure p derive the equation expressing the 
relation between the volume and the pressure. Plot for pressures from 
J atmosphere to 10 atmospheres. 

2. Given an oak beam 10 ft. long of such dimensions that it supports 
2 tons at its midpoint when resting at each end upon a support. If 
the weight w such a beam will support varies inversely as its length I 
derive the equation expressing the relation between w and L Plot for 
values of I from 1 ft. to 20 ft. 



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152 ANALYTIC GEOMETRY [§119 

3. Find the locus of the center of a circle tangent externally to two 
given circles. 

4. Find the locus of the center of a circle haying one of two given 
circles tangent to it internally and the other tangent to it externally. 

6. The base of a triangle is fixed, and the difference of the angles at 
the base is $*\ Find the locus of the vertex opposite the base. 

6. Three instruments for recording sound are located at three points, 
A f B, and C, in a straight line. From A to B is 300 ft. and from B to 
C is 500 ft. A sound, such as the report of a cannon, is recorded at B 
0.05 second after it is recorded at A, and at C 0.35 second after it is 
recorded at B. Find the location of the source of the sound in distance 
and direction from the point midway between A and B. 

Suggestion. — Choose the origin of coordinates at the point midway 
between A and B. Derive the equations of the hyperbolas and solve as 
simultaneous equations. Note that only one branch of each hyperbola 
is possible. 

The equations of the hyperbolas will be found to be 
27.7s» - y* - 21716, 
and' {x - 400)* - 1.6y* = 38416. 

The solution of these equations gives 

x = - 70 and y - 337.7. 

GENERAL EXERCISES 

1. Find the semi-axes, the eccentricity, and the coordinates of the 
foci of the hyperbola 2a? 1 — Zy* — 12. Also find the equation of the 
hyperbola that is conjugate with this. 

2. Find the coordinates of the points of intersection of the hyperbola 
2x* - Zy 1 = 12 and the circle x % + y 1 = 16. 

3. Find the semi-axes, coordinates of foci, eccentricity, and equations 
of directrices of the hyperbola 9x* — 4y* — 54s + 16y + 209 « 0. 

4. Show that the following equation represents two straight lines 
parallel respectively to the codrdinate axes: 12xy + 8z — 27y — 18 = 0. 

Transform the following equations as indicated, illustrating each 
by a drawing: 

5. a* - lOxy +y*+x + y + l=0ti> 32s* - 48y* - 9. 

6. x* - 2xy - y* - 2 - to x* - y* + V^ - 0. 

7. Find the equation of the locus of a point that moves so that the 
difference of its distances from ( —4, 2) and (4, 2) is always equal to 8. 

x* y 2 

8. Given the hyperbola — = 1, find the coordinates of the 

25 16 

point on the hyperbola, with abscissa double the ordinate. 



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§119] THE HYPERBOLA AND CERTAIN FORMS 153 



9. Find the distances from the foci of the hyperbola -— — -— — 1 to a 

25 16 

point on the hyperbola, with abscissa 10. 

10. Find the equation of an hyperbola whose axes are parallel respec- 
tively to the coordinate axes and which passes through the points (0, 0), 
(1,1), (-2, -1), and (-2, 2). 

11. The lines x — 2y — and x + 2y — are the asymptotes of an 
hyperbola that passes through the point (—5, 3). Find its equation. 

12. Prove that for all values of a the point (a sec a, b tan a) is on the 

x s y* 
hyperbola — — — = 1. 
a* b* 

13. Prove that sec a is the eccentricity of an hyperbola with asymp- 
totes including an angle 2a. 

14. Prove that the portion of an asymptote of an hyperbola, which 
is intercepted between the directrices is equal to the transverse axis. 



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CHAPTER IX 
OTHER LOCI AND EQUATIONS 

120. General statement — In the previous chapters, for 
the most part, equations of the first and second degree in two 
variables, and their loci are considered. In the present 
chapter a consideration will be made of other equations also 
and their loci, where they are of importance in the study of 
more advanced mathematics, or are of use in immediate 
applications to science and engineering. 

Such equations and loci are of infinite variety and form. 
They may be divided into two classes, (1) algebraic' and 
(2) transcendental. 

Algebraic curves the degree of whose equations is higher than 
the second, and all transcendental curves that he wholly in a 
plane, are often called higher plane curves. 

In Cartesian coordinates an equation that can be expressed 
in a finite number of terms of the form Qx n y m , in which the 
variables are affected by constant exponents and Q is a con- 
stant, is called algebraic, all others are called transcendental. 

121. Summary for second degree equations. — The most 
general equation of the second degree in two variables may be 
written in the form 

Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. 
Theorem. — In rectangular coordinates, the equation of the 
second degree in two variables represents a conic section. 

To prove this it is only necessary to show that, by a suitable 
change of the coordinate axes, the equation reduces to a form 
already discussed. 

Given Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. 

154 



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§121) OTHER LOCI AND EQUATIONS 155 

From [13] substituting x = x' cos <p — y' sin <p f 
and y = x 1 sin ^> + y' cos ^>, 

A(a' cos <p— y'8in<p) 2 +B(x' cos<p—y' 8in<p) (x' sin ^>+y , cos^>) + 
C(z' sin ^> + y' cos ^) 2 + D(x' cos <p — y' sin ^) + 

2?(z' sin ^> + y 1 cos ^) + F = 0. 
Expanding and collecting terms, 

x' 2 (A cos 2 <p + B sin <p cos ^ + C sin 2 ^>) + 
x'y'(—2A sin ^cos^ + 2C sin <p cos ^ — B sin 2 ^ + B cos 2 <p) + 
. j/ ,2 (A sin 2 <p — B Em <p cos <p + C cos 2 ^) + 
x'(D coa <p + E sin ^>) + y'{E cos ^ — D sin ^) + F = 0. 
In this equation the x'y 1 term will vanish if 

— 2A sin <p cos <p + 2C sin ^cos <p — B sin 2 ^ + B cos 2 ^ = 0. 
Or if B(cos 2 <p — sin 2 <p) = (A — C)2 sin ^ cos ^. 

By trigonometry, this becomes B cos 2<p = (A — C) sin 2^>. 

B 

[39] . • . tan 2* = j^zTq' 

Since the tangent of an angle may have any value from 

— a> to + °° > it is always possible to rotate the coordinate 
axes through such an angle that the zV-term will vanish. 

Further, since the smallest positive value of 2<p is less than 
180°, <p is an acute angle. This value of <p can always be 
chosen for the rotation. 

The general equation then reduces to the form 

A'x 2 + C'y 2 + D'x + E'y + F' = 0. 

From the considerations of the previous chapters, this 
equation represents one of the conic sections as follows: 

(1) A circle if A' = C". 

(2) A parabola if A' ^ 0, E' ^ 0, and C" = 0, 

or if C" ?* 0, D' * 0, and A' = 0. 

(3) An ellipse if A' and C are of like signs and unequal. 

(4) An hyperbola if A' and C have unlike signs. 

Theorem. — The general equation o f the second degre e in x 
and y, Ax 2 + Bxy + Cj/ 2 _dr, Zta ,+ % $ X^ , mpni mtfaT 



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4A- 



t i 156 ANALYTIC GEOMETRY (§12i 

p arabola? im -^Hfggg^or on hyperbola according a$ B 2 — 44. C 
fYpmfo awe, t'g fcafl lftan sero r or & flflflE dF %e*i g^rvT 
Proo/. — Using the values of A', £', and C", 

ii' = ii cos 2 ^ + Bsin^cos^ + C sin 2 <p, (1) 

5' = B(cos 2 <p — sin 2 ^) — (A — C)2 sin ^ cos <p, (2) 
C — A sin 2 ^ — B sin ^ cos tp + C cos 2 <p. (3) 

Adding (1) and (3), 

A' + C" = A + C. (4) 

Subtracting (3) from (1), 

A' - C" = (ii - C) cos 2*> + £sin2*>. (5) 

Squaring (2) and (5) and adding, 

B' 2 + (A' - CO 2 = B* + U - C) 2 . (6) 

Squaring (4) and subtracting from (6), 

£' 2 - ±A'C = B 2 - 4AC. (7) 

But, if ^ is chosen so that tan 2<p = a _ n 1 &' ~ 0- 

Hence B 2 - 4AC = -4A'C. 

From this it follows that ' > 

(a) B 2 - 4AC - if either A' - or C = 0, \ 
(6) B 2 - 4AC<0 if A' and C" have like signs, 
(c) B 2 - 4AO0 if A' and C" have unlike signs. J r v " 
These are respectively the conditions necessary for a parabola, 
an ellipse, or an hyperbola. J M - *\ r v* 

Example,— -Given s 2 + 24a# - 0y* + 4s + 48y + 34 - 0. (1) Deter- 
mine whether it represents a parabola, an ellipse, or an hyperbola; (2) 
transform so as to free of the xy-term; (3) reduce to the standard form; 
(4) plot and show the three sets of axes. 

Solution.— (1) B» - 4AC = 24* - 4.1(-6) - 600. 
\ Therefore the equation represents an hyperbola. 

\ (2) tan 2<p - j^Tc " *+' cos 2<p " A * 
sin *> - yt(l - cos gg = V j(l - JV) - f. 
cos * - VJ(1 + cos 2*) - Vi(l + A) - f • 
; Then the formulas [18] become x — |x' — f y 7 , and y — fa/ 4- f y'. 



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»22j: 



OTHER LOCI AND EQUATIONS 



157 



Substituting in the equation, {& - fyO* + 24(f* / - &)(%* + fyO 
- 6(4^ +t/)* +4(^ - f^) +48(Js' + !•) + 34 - 0. 
Simplifying, 10s" - 15Y» + 32^ + 36^ + 34 - 0. 
(3) Putting x 1 - x" + h and y* - y" -f fc, and simplifying, 
10s"» - 15y"» + (20fc + 32)*" - (30fc - 36)y" + lOfc* - 15fc* + 

Z2h + 36fc + 34 - 0. 
Equating coefficients of x" and y" to 0, and solving, 

20h + 32 - 0, 30fc - 36 - 0. .*. fc - -|, and k - f . 
Substituting these values and simplifying, 



2*"» 



3y"* + 6«0, or Y»^-= 1. 



This is an hyperbola with its center at the origin and its transverse 
axis along the y"-axis. 

(4) The three sets of coordi- 
nate axes and the curve are as 
shown in Fig. 108. 

Remark. — The values of 
h and k could have been 
found by completing the 
squares in x and y. In 
the solution given above, 
the rotation of axes was 
made first; but the work 
would have been shortened 
somewhat if the axes had 
been translated first. 

122. Suggestions for simplifying second degree equa- 
tions. — If the equation is that of an ellipse or hyperbola, 
first translate the axes to remove the terms of the first degree, 
and then rotate the axes to remove the xy~term. 

If the equation is that of a parabola, first rotate the axes to 
remove the xy-term, and then translate to remove the constant 
term and one of the terms of first degree. 

It may-he. that the locus is a perrrtTGiat it" is composed of 
straight lines, or is imaginary. These forms are often called 
degenerate forms and are best discovered from the simplified 
E quation. — " ..- - 




Fio. 108. 



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158 



ANALYTIC GEOMETRY 



(§123 



EXERCISES 

Test each of the following equations as to whether it is a parabola, an 
ellipse, or an hyperbola. Simplify each and plot showing all sets of , 
coordinate axes. > ,\., f[ A v t , ^ }/ 



/ 



/ 1. 6s* + 24sy - y* + 5% - 55 - 0. * ' 

2. 25s* - Uxy + 25y* + 142s - 178j/ + 121 = 0. 

-\ 3. s* + sj/ + y* - 3y + 6 = 0.(2^ 

/ 4. 32s* - 4Sxy + lSy* + 35s - 120y + 200 - 0. *V ; ' 

I 6. 13s* - 6\/5sy + 7y* - 64 - 0. , t 

6. s 2 - 2y/Zxy + 3y* - 6\/3z - fy - 0. " 

^ 7- 2s* + 6sy + 10y* - 2s - 6y + 19 - 0. 

8. 6s* + 13sy + 6y* - 8s - 7y + 2 = 0» 

9. 4s* + 4sy + y* + 4s - 3y + 4 - 0. 

10. 9s* - 12sy + 4#* - 20s - 30y - 50 = 0. 

ALGEBRAIC EQUATIONS 

123. Parabolic type. — Equations of the form y = oaf, 
where a is a constant and n is positive, are said to be of the 
parabolic type. 
I (1) When n = 2, y = ox 2 . The locus is the ordinary- 
parabola with its axis on the j/-axis, and has 
already been discussed. 

(2) When n = 3, y = ox 8 . The locus is 
called the cubical parabola. It has the form 
shown in Fig. 109, for a — 1. 

Discussion. — When s = 0, y = 0, and the curve 
passes through the origin. It is not symmetrical 
with respect to either coordinate axis, but is sym- 
metrical with respect to the origin. Why? 

For any positive value of s, y is positive; and for 
any negative value of s, y is negative. Hence the 
curve lies wholly in the first and third quadrants. 

This information together with a few points makes 
Fig. 109. it possible to sketch the curve with considerable 

accuracy. 

(3) When n = I, y = axi. The locus is called the semi- 
cubical parabola. It has the form shown in Fig. 110, for 
o - 1. 




*»x 



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§124] 



OTHER LOCI AND EQUATIONS 



159 



Discussion. — When x = 0, y = 0, and the curve passes through the 
origin. Writing y = x* in the form y* — x z , it is seen that the curve 
is symmetrical with respect to the z-axis. 

For any positive value of x, y has two values 
numerically equal but opposite in sign. For any 
negative value of x, y is imaginary. Hence the curve 
lies wholly in the first and fourth quadrants. 

124. Hyperbolic type. — Equations of the 
form y = ax n , where a is a constant and n is 
negative, are said to be of the hyperbolic 
type. 

(1) When n = — 1, y = aar 1 , or xy = a. 
The locus is the ordinary equilateral hyperbola 
lying in the first and third quadrants. 

(2) When n = —2, y = ax~ 2 y or x*y ^ a. 
The locus has the form shown in Fig. Ill, for 
a = 1. Fiq. no. 

Discussion.— No finite value of x will make y = 0, and no finite 
value of y will make x = 0. Hence the curve does not meet either of 
the coordinate axes. 





*-x 



=*•* 



Fia. 111. 



Fiq. 112. 



Since x is affected only by an even exponent and y only by an odd 
exponent, the curve is symmetric only to the j/-axis. 

For all positive finite values of y, x has two finite values equal numeri- 
cally but opposite in sign. For all negative values of y, x is imaginary. 
As y becomes large positively, x approaches zero both from the positive 



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160 



ANALYTIC GEOMETRY 



[§125 



and the negative side. As x becomes large either positively or negatively, 
y approaches zero from the positive side. 

The curve lies in the first and second quadrants, and is asymptotic 
to both coordinate axes. 

(3) When n = — f, y = ax~l, or xty = a. The locus has 
the fdrm shown in Fig. 112, for a = 1. 
The discussion- is left as an exercise. 

EXERCISES 

Plot each group of the following equations upon the same set of 
coordinate axes, by first discussing the equation and then finding a 
few points. 

LY I 1. (1) V - x\ (2) y - x\ (3) y « x\ 

~ ^ (2) !/-*«, (3)y-x*. 

(2) y=*«, (3)y«*«. 
(2)y=xl 
(2)y-ar». 
(2)y=ar«. 
(2)y = aT*. 
(2)y-ar«. 

9. If p is the pressure and * the absolute 
*■ JC temperature of a gas in adiabatic expan- 

7 

sion, p = fct 7 "" 1 , where fc is a constant and 
y — 1.41 for air. If p « 2700 when 
t = 300, find fc, and plot the equation for 
values of t from 200 to 400. 

10. In a mixture in a gas engine expand- 
ing without gain or loss of heat, it is found 
that the law of expansion is given by the 
equation pv 1 * 91 « c. Given that p = 188.2 
when v =* 11, find the value of the constant 
c, and plot the curve of the equation using 

Consider values of v from 10 to 25. 




1. 


(i) v 


= *», 


2. 


(1) V 


= *, 


S. 


(1) V 


-»», 


4. 


(i) y 


-*», 


6. 


(1) V 


-*-', 


6. 


Wv 


= *-«, 


7. 


(1) V 


-=r*, 


8. 


(i) y 


-*-«, 



Fig. 113. 
this value of the constant. 



125. The cissoid of Diodes.— In Fig. 113, OT is the 

diameter of a fixed circle. At fa tangent is drawn, while 
about a secant revolves meeting the tangent in Q and the 
circle in R. The point P on the line OQ is taken so that 
OP = RQ. The locus of the point P is the cissoid of Diodes, 



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§126] OTHER LOCI AND EQUATIONS 161 

To derive the equation of the cissoid, choose as origin and 
the x-axis along 07V- ^ 

Draw MP and NR perpendicular to OT. 

Denote the coordinates of P by (x, y). 

Let the radius of the circle be a. <- > \ 

From the definition of the cissoid, OP = RQ. 

And evidently OM = NT. 

But OM = x, hence NT - x and CW = 2a - x. 

Also NR is a mean proportional between ON and iVT. 

Hence NR = Vs(2a — x). ' v 

By similar triangles, OM : ON = MP : NR. 

Substituting values, x : 2a — x = y : Vs(2a — £)• 

x 3 . ' 

From this y 2 = ^ — ~ — > the equation required. 

The curve may be plotted from the definition given above, 
or from the equation. 

Note. — By means of the cissoid the problem of the dupli- 
cation of the cube can be solved. This problem, to find a cube 
that is double a given cube, was one of the famous problems of 
antiquity. 

126. Other algebraic equations. — An unlimited number of 
definitions of loci could be given that would result in algebraic 
equations. There are many such curves that are of more or 
less historical importance as well as of value in mathematics 
and other sciences. Also there are an unlimited number of 
algebraic equations that may be discussed and their curves 
plotted. It should be remembered that an algebraic equation 
as truly defines a locus in terms of rectangular coordinates, 
as does the definition of the preceding article define the 

cissoid. 

8a' 



Example 1. — Discuss and plot the equation y = 



x* +4a* 



Intercepts, — When x = 0, y — 2a. The curve does not meet the 
rr-axis, since no finite value of x will make y = 0. 

Symmetry. — Since only an even power of x occurs the curve is sym- 
metrical with respect to the y-axis. 
11 



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162 



ANALYTIC GEOMETRY 



[§126 



Extent. — If a is a positive number, y is positive for all real values of 
x, and the largest value of y is 2a when x = 0. As x becomes very 
large in absolute value, y becomes very small but always positive. 
Hence the curve is asymptotic t o the s- axis in both directions. 

Or, solving for x, x » ±2a *l/ • 

Hence x is imaginary when <o. 

This is true when y<0 or when y>2a. 

Hence the curve lies in the first and second quadrants, is symmetrical 
with respect to the y-axis, and lies between the x-axis and the line y = 2a. 

Points on the curve and in the first quadrant can be found by choosing 
positive values for x. 



X 





a 


2a 


3a 


4a 


6a 


y 


2a 


i« 


a 


*» 


i« 


& 



The curve is as shown in Fig. 114, and is known as the witch of Agnesi. 

Y 



« 



Fig. 114. 

EXERCISES 
Discuss the equations and plot the curves in exercises 1-16. 




1. y - 

2. y = 



8 



x -2 

x + 3 



9. xV - (V + 2) 8 (16 - y«). 
(Example of Conchoid of Nicho 



x -3 

3. y = x(x — l)(x — 2). 

4. y* = (x - l)(s - 4)(s - 6). 
x +3 

(s - 2)(s + 1)* 
(a? - 4)(s + 3) 
(x + 1)<* - 2)' 

7. x\ + y* = a*. 

8. si 4- y* = <**• 



«. y 
6. y 



10. xy* = (* - a) 8 (2a - x). 

11. 9y* = (s + 7)(3+4)«. 

12. y = x 8 + x - 3. 

13. y - *'+ 6s* + 10s - 2. 

14. y = s< - lOz* - 4s + 8. 

16 - * = FT* - 

16. «*)■ + (Mi - I- 



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§127] OTHER LOCI AND EQUATIONS 163 

• 

17. Two fixed points F f and F are 2a units apart. Choose the 
origin at the center of the line joining F' and F, and the z-axis along this 
line. Find the equations of the loci of the point P(x, y) when 

FP 

(1) pfp — a constant not unity, 

(2) FP + F'P - a constant, 

(3) FP - F'P = a constant, 

(4) FP X F'P - a constant, k. 

In (4) the locus is called a Cassinian oval, and its equation is 
(** + y*)* - 2a 8 (s* - y*) - h* - a*. 

18. Sketch the loci of (4) of the preceding exercise when a — 1, and 
k has successively the values 0, 1, and 2. 

10. Write the equation of (4) of exercise 17 when k — a f , and plot 
the curve. This curve is called a lemniscate. 

20. Express the lemniscate in polar coordinates, using the positive 
part of the z-axis as polar axis. 

21. A uniform beam of length I, fixed in position by being held at 
one end, supports a weight at the other end. The deflection y at any 
distance x from the fixed end is given by the equation y — k(\lx* — Ja? 8 ). 
Find k for a beam 12 ft. long if the weight deflects the outer end 18 
in., and plot a curve showing the shape of the beam for its entire length. 
Choose the fixed end as the origin and consider y positive when measured 
downward. 

TRANSCENDENTAL EQUATIONS 

127. Exponential equations. — An equation of the form 
y = b x , where 6 is any positive constant, is called an expo- 
nential equation. If the exponent is fractional and involves 
even roots of 6, only the positive values of these roots are 
used. 

Example 1. — Discuss the equation y = b* when 6>1. Plot the curve 
when b = 1.5. 

Intercepts. — When x — 0, y = 6° » 1. This shows that the curve 
passes through the point (0, 1) for any value of b. 

If y = 0, b* = 0, which is impossible for any finite value of x. This 
shows that the curve neither meets nor crosses the rc-axis. However, 
for sufficiently large negative values of x, the value of b* can be made 
to become as near zero as desired. The curve is then asymptotic to 
the a>axis in the negative direction. 

Symmetry. — Since changing sto — 3 or y to — y changes the equation, 
the curve is not symmetrical with respect to either coordinate axis. 



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164 



ANALYTIC GEOMETRY 



[§128 



Extent.— Since no integral value of a; can make y negative, and since 

only positive values of b* are to be 
taken when 2 is a fraction, the curve is 
wholly above the z-axis. 

Further, since y is not imaginary for 
any value of x, and increases as z in- 
creases, the curve lies in the first and 
second quadrants, exists for all values 
of x, and continually rises from left to 
right. 
x Plotting. — The curve of y m 1.5* can 
be plotted as accurately as desired by 
finding points. Taking logarithms of 
Fig. 115. both sides of the equation, 

log y = x log 1.5 = 0.1761a;. 
The following points are readily found, and the curve is as shown 









































































































































(1) 








(2) 










































































































































in(l) 


of Fig. 


115. 
















X 


-3 


-2 


-1 





1 


2 


3 


4 


5 


logy 


1.4717 


1.6478 


T.8239 





0.1761 


0.3522 


0.5283 


0.7044 


0.8805 


y 


0.296 


0.444 


0.667 


1 


1.5 


2.25 


3.375 


5.063 


7.595 



Example 2. — Discuss the equation y = b* when 6<1. Plot the curve 
when b * J- 

The discussion is similar to that of example 1. It is to be noted 
that y decreases as x increases, and the curve is asymptotic to the z-axis 
in the positive direction. 

Plotting. — Points for plotting y = (J)* are found and the curve is 
as shown in (2) of Fig. 115. 



X 


-5 


-4 


-3 


-2 


-1 





1 


2 


3 


4 


v 


32 


16 


8 


4 


2 


1 


0.5 


0.25 


0.125 


0.0625 



128. Applications. — The most important case of the expo- 
nential equation is the case where the base is e, which is 
the base of the natural system of logarithms and equals 
2.71828 • • < , It usually occurs in the form y = ae**, where 



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§129] OTHER LOCI AND EQUATIONS 165 

a and k are constants, that may be determined in particular 
applications. This function is often called the "law of organic 
growth," or the "compound interest law," and is a function 
where the rate of increase or decrease at any instant is directly 
proportional to the value of the function at that instant. 
Just what the applications are cannot well be shown here, 
but the following uses are suggestive : 

(1) To express the pressure of the atmosphere at any height. 

(2) In physics and electricity, it is used in considering 
damped vibrations. 

(3) In medicine and surgery, to express the progress of the 
healing of a wound. 

(4) In biology, to determine the number of bacteria in a 
culture at any given time. 

(5) In chemistry, to express the progress of a chemical action. 

(6) In mechanics, in connection with the slipping of a belt 
on a pulley. 

Numerous applications will be discovered by the student 
as he progresses in his studies. 

Because of its frequent occurrence in problems involving 
conditions in nature, the base e is sometimes called ' ' a constant 
of nature." 

129. Logarithmic equations. — The logarithmic equation is of 
the form y = logi, x, where 6 is a positive number different from 1. 

By the definition of the logarithm of a number, the equation 
y = log& x can be written in the exponential form x = b v . 
This is the same as the equation of article 127 with x and y 
interchanged. 

It is evident then that the discussion of the logarithmic 
equation y = log& x follows that of the exponential equation 
jc = b v } and gives the following when b > 1 : 

The x-intercept is at the point (1, 0). 

There is no ^/-intercept for as x approaches 0, y becomes 
— » , that is, the curve is asymptotic to the y-axis in the 
negative direction. 



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166 



ANALYTIC GEOMETRY 



[§129 



The curve is not symmetrical with respect to either axis. 

When x > 1, y >0, and as x becomes » , y becomes «> also. 

When x < 1, y <0, and there is no value of y which will make 
x negative. 

Example 1. — Plot the curve of y = logio x. 

The following points are found, and the curve is (1) of Fig. 116. The 
unit on the y-axis is taken twice that on the x-axis. 



X 


0.001 


0.01 


0.1 


0.5 


1 


2 


3 


4 


5 


7 


10 


15 


50 


100 


V 


-3 


-2 


-1 


-0.301 





0.301 


0.477 


0.602 


0.699 


0.845 


1 


1.176 


1.699 


2 




Fig. 116. 

Example 2. — Plot the curve of y = log 2 x. 

The points can be readily found from the relation 

\ogiX = . s X log™ x = 3.322 log 10 x. 

lOgio A 

The curve is (2) of Fig. 116. 



X 


0.001 


0.01 


0.1 


0.5 


1 


2 


3 


4 


5 


7 


10 


15 


100 


V 


-9.97 


-6.64 


-3.32 


-1 





1 


1.58 


2 


2.32 


2.81 


3.32 


3.91 


6.64 



EXERCISES 

1. Plot the following exponential equations: 

(1) y =- 2*. (2) y = 3*. (3) y = e*> where e - 2.718. 

(4) y = (0.75)*. (5) y - (0.4)*. 

2. Discuss the effect upon the curve of y « 6* when 6< 1 and increases 
tol. 



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§130] OTHER LOCI AND EQUATIONS 167 

8*. Discuss the effect upon the curve of y — b* when b> 1 and increases 
from 1. 

4. Plot the curves of the following: 

(1) y = l-». (2) y - 2~». (3) y - *-, x>0. (4) y - «r*. 

5. Plot the curves of the following: 

(1) y * log, x f where e - 2.718. (2) y - log* 2. (3) y - 2 log,*. 

6. Discuss and plot the curve of y * — « — " 

Suggestion. — First plot yi » Je* and y% « Je~*. Then plot 

6* -\-e~ 9 
y » — 5 — by adding the ordinates j/i and yi to find y for the 

dfferent values of x. 

7. Discuss and plot the curve of y » Ja(«o -f- «~ a ). This is the 
equation of the catenary, the curve assumed by a flexible cord sus- 
pended between two points. 

8. A wire, weighing 0.2 lb. per foot, is suspended from two points in 
a horizontal line 50 ft. apart. The horizontal tension at each end 
is 10 lb. Plot the catenary formed by the wire. The constant a 

in the formula, y » Ja(e<* •+■ e~ a), is found by dividing the horizontal 
tension by the weight per unit length of the wire. 

9. Plot the curve of i = ber**, where i and t are the variables. Choose 
b - 1.5 and a » 0.4. 

10. If a body is heated to a temperature 7\ above the surrounding 
bodies, and suspended in air, its excess of temperature T above the 
surrounding bodies at any time, t seconds thereafter, is given by Newton's 
law of cooling expressed by the equation T — Tie""*, where a is a 
constant that can be determined by experiment. Given 7\ =■ 20 and 
a =- 0.014, plot a curve showing the temperature at any time t up to 
100 seconds. 

11. The dying away of the current on the sudden removal of the 

electro-motive force from a circuit containing resistance and self-induc- 

_ ?* 
tion, is -expressed by the equation, % = If L, where i is the current 

at any time, t seconds, after the e.m.f. is removed, R is the resistance, 

and L the coefficient of self-induction. Plot a curve to show the current 

at any time from t = to t = 0.2, if / » 10 amperes, #=0.1 ohm, 

and L = 0.01 henry. 

TRIGONOMETRIC EQUATIONS 

130. The sine curve. — Discuss the equation y = sin x, 
and plot the curve. 



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168 



ANALYTIC GEOMETRY 



[§131 



Intercepts. — When x = 0, y = 0. Hence the curve passes 
through the origin. When y = 0, sin x = 0, and x = mr 
radians, where n is any integer either positive or negative. 

Symmetry. — Putting — y for y or — x for x, changes the 
equation. Hence the curve is not symmetrical with respect 
to either axis. But putting — y for y and —x for x, does not 
change the equation. Hence the curve is symmetrical with 
respect to the origin. 

Extent. — Since there is a sine of any angle, the curve extends 
indefinitely in both the positive and negative directions. 

Since the sine of an angle is not greater than 1 nor less than 
— 1, the curve does not extend above the line y = 1 nor below 
the line y = — 1. 

Plotting. — Any length can be chosen as a unit on the coordi- 
nate axes. What may be called the proper sine curve is 




Fig. 117. 



plotted by choosing as a unit on the j/-axis the same length 
that is chosen to represent one radian on the x-axis. The 
curve is shown in Fig. 117. 



X 






.5 


.707 


** 


i* 


br 


** 


** 


X 


br 


J* 


** 


br 


*T 


br 


¥* 


2x 


y 


.866 


1 


.866 


.707 


.5 





-.5 


-.707 


-.866 


-1 


-.866 


-.707 


-.5 






From 2t radians to 47r radians or from — 2t radians to 0, 
these values repeat. They also repeat for each interval of 
2w radians in both directions. 

131. Periodic functions. — A curve that repeats in form as 
illustrated by the sine curve is called a periodic curve. The 
function that gives rise to a periodic curve is balled a periodic 
function. The least repeating part of a periodic curve is 



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§132] 



OTHER LOCI AND EQUATIONS 



169 



H* 



K* 




2T 



+X 



Fig. 118. 



sin 2x. 



called a cycle of the curve. The change in the value of the 
variable necessary for a cycle is called the period of the func- 
tion. The greatest absolute value of the ordinates of a periodic 
function is called the amplitude of the function. 

In engineering and other practical applications of mathe- 
matics, there are many phe- 
nomena that repeat. It is for 
this reason that the periodic 
functions are of great import- 
ance. By a suitable choice of 
periodic functions almost any 
periodic phenomenon can be 
represented by a function. 

132. Period and amplitude of a function. 

Example 1. — Find the period of sin nx, and plot y 

Since, in finding the value of sin nx, the angle x is multiplied by n 

before finding the sine, the period is — • . 

The curve for y = sin 2x is shown in Fig. 118. The period of the 
function is ir radians, and there are two cycles of the curve in 2x radians. 

Definition. — The number n 
in sin nx is called the period- 
icity factor. 

Example 2. — Find the amplitude 
of b sin x, and plot y — 2 sin x. 

Since, in finding the value of 
b sin x, sin x is found and then 
multiplied by b, the amplitude of 
the function is b, for the greatest 
value of sin x is 1. 

The curve for y — 2 sin x is shown 
in Fig. 119. The amplitude is 2. 

Definition. — The number 6 in 6 sin x is sometimes called 
the amplitude factor. 

By a proper choice of a periodicity factor and an amplitude 
factor a function of any amplitude and any period desired can 
be found. 




Fig. 119. 



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170 



ANALYTIC GEOMETRY 



[§133 



133. Projection of a point having uniform circular motion. 
Simple harmonic motion. 

Example 1. — A point P, Fig. 120, moves around a vertical circle of 
radius 3 inches in a counter-clockwise direction. It starts with the 
point at A and moves with an angular velocity of 1 revolution in 10 
seconds. Plot a curve showing the distance the projection of P on the 
vertical diameter is from at any time *, and find its equation. 

Plotting. — Let OP be any position of the radius drawn to the moving 
point. OP starts from the position OA and at the end of 1 second 







Pi tf . J 


P. 


Y 




















P^ 


A 






Sf" 


rT 


7 






















r 




-Jvj 






?JL_^_^ 


H— 


















f 




























r^M 


i. 




o 




p 


Y>* ! 


2 
linn 


in t 


> ■ 

MOO 


1 
II 


^ 1 


i ' 


' 1 


i 1 


k 


L__. 






—A 


1... 
















— ) 


* 


s 




22 


s'. 


9 


L__. 












fc-"i 


/ 






p 




i 


\ 























Fig. 120. 

is in the position OP h having turned through an angle of 36° =» 0.6283 
radians. At the end of 2 seconds it has turned to OP*, through an 
angle of 72° — 1.2566 radians, and so on to the positions OP it OPi t - • *, 
OP 10 . 

The points Ni, N%, • • • are the projections of Pi, Pj, • • • respect- 
ively, on the vertical diameter. 

Produce the horizontal diameter OA through A, and lay off the seconds 
on this to some scale, taking the origin at A, 

For each second plot a point whose ordinate is the corresponding 
distance of N from O. These points determine a curve of which any 
ordinate y is the distance from the center O of the projection of P 
upon the vertical diameter at the time t represented by the abscissa 
of the point. 

It is evident that for the second and each successive revolution, the 
curve repeats, that is, it is a periodic curve. 

Since the radius OP turns through 0.6283 radians per second, angle 
AOP - 0.6283* radians, and ON » OP- sin 0.6283*. Or y » 3 sin 0.6283*, 
the equation of the curve. 



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§133] 



OTHER LOCI AND EQUATIONS 



171 



In general, then, it is readily seen that if a straight line of 
length r starts in a horizontal position when time, t = 0, and 
revolves in a vertical plane around one end at a uniform 
angular velocity o> per unit of time, the projection y of the 
moving end upon a vertical straight line has a motion 
represented by the equation 

y = r sin <at. 

Similarly, the projection of the moving point upon the hori- 
zontal is given by the ordinates of the curve whose equation is 

y = r cos <at. 

The motion of the point N is a simple harmonic motion. 

If the time is counted from some other instant than that 
from which the above is counted, then the motion is 
represented by 

y =* r sin (<at + a), 
where a is the angle that OP makes with the line OA at the 
instant from which t is counted. As an illustration of this 
consider the following: 




Fig. 121. 

Example 2. — A crank OP, Fig. 121, of length 2 ft. starts from a posi- 
tion making an angle a = 40° = |r radians with the horizontal hne 
OA when t = 0. It rotates in the positive direction at the rate of 2 
revolutions per second. Plot the curve showing the projection of P 
upon a vertical diameter, and write the equation. 

Plotting. — The axes are chosen as before, and points are found for each 
0.05 second. The curve is as shown in Fig. 121. 

The equation is y — 2 sin (4rJ +#*•). 



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X72 ANALYTIC GEOMETRY [§134 

Definitions.* — The number of cycles of a periodic curve in a 
Unit of time is called the frequency. 
It is evident that 

J — yF> 

where / is the frequency and T is the period. 

In y = r sin (at + a), f = jr- and T 7 = — 

^71* to 

» 

The angle a is called the angle of lag. 

134. Other applications of periodic functions. — The illustra- 
tions already given are by no means the only uses of periodic 
functions. Many uses occur in connection with sound, light, 
and electricity. Periodic curves are traced mechanically on 
smoked glass in experiments in sound and electricity. Such 
curves are also traced by instruments for recording heartbeats, 
breathing movements, and tides. 

Any periodic motion can be represented exactly, or can be 
closely approximated, by functions involving sines and cosines. 

135. Exponential and periodic functions combined. — The" 
curve represented by the equation, y = be ax sin (nx + a), is 
important in the theory of alternating currents, in representing 
the oscillations of a stiff spring, the damped oscillations of a 
galvanometer needle, or the oscillations of a disk suspended in 
a liquid, such as is used to compare the viscosities of different 
liquids. 

The curve is most readily plotted by first plotting the curves 
represented by the exponential function and the periodic 
function separately on the same set of axes, and then finding 
the ordinates for various values of x by multiplying together 
the ordinates for these values of x in the exponential and 
periodic functions. 

It will be noted that the curve is periodic, and that the 
amplitude of the successive waves gets less and less while the 
wave length remains the same. 



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§136] OTHER LOCI AND EQUATIONS 173 

Example. — Plot the curve showing the values of y for any value of x 
from x=— }*■ to a; = 2r f or the equation, y = e~°- 6 * sin (2x + Jx). 

The curve is readily plotted by first plotting j/i ■» e~°- oto and 
y s = sin (2a? + Jx), and then finding various values of y from the 
relation y = j/it/j. In Fig. 122, (1) is the exponential curve, (2) the 
sine curve, and (3) the final curve. Note that (3) and (2) intersect 
the x-axis at the same points. 



Fig. 122, 



EXERCISES 



1. Plot y = sin x t using several different lengths on the z-axis as units. 

2. Discuss and plot y = cos x. Give its period. 

3. Discuss, and plot y = tan x, and y = cot x on the same set of 
axes. Give the period of each. 

4. Plot y = sin x + cos x. 

Suggestion. — Plot f/i = sin x and y% — cos x on the same set of axes. 
Then find y from y = y\ + #a> by adding the ordinates for various 
values of x. 

6. Plot y =* sin 1 x and y = cos 2 x on the same set of axes. 

6. Plot y = sin"" 1 x and ?/ = cos" 1 re. 

7. Plot y = sec a; and y = esc s, and give the period of each. 

8. Plot y = sin \x } y = sins, y = sin2x, and y = sin fa; on the same 
set of axes. 

9. Plot y = i sin s, y = sin x, y = 2 sin x t and y = f sin x on the 
same set of axes. 



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174 ANALYTIC GEOMETRY [§136 

10. Plot y = sin 2x + 2 cos x, and give the period. 

11. Plot y — sin 3 + x. Is this periodic? 

12. A crank 18 in. long starts from a horizontal position and rotates 
in the positive direction in a vertical plane at the rate of !*■ radians 
per second. The projection of the moving end of the crank upon a 
vertical line oscillates with a simple harmonic motion. Construct a 
curve that represents this motion, and write its equation. 

13. A crank 8 in. long starts from a position making an angle of 55° 
with the horizontal, and rotates in a vertical plane in the positive direc- 
tion at the rate of one revolution in 3 seconds. Construct a curve showing 
the projection of the moving end of the crank in a vertical line. Write 
the equation of the curve and give the period and the frequency. 

14. Plot the curves that represent the following motions: 

(1) y = 12 sin (1.88* + 0.44), (2) y - 2.5 sin (Jx* + T i*x). Give the 
period and frequency of each. 

16. Plot y = rsin fat and y «■ r sin (Jx* + W" on the same set 
of axes. Notice that the highest points on each are separated by the 
constant angle Jx. Such curves are said to be out of phase. The 
difference in phase is stated in time or as an angle. In the latter case 
it is called the phase angle. 

16. Plot y — r sin Jirf, y = r sin {\xt — Jx), and y = r cos Jxi all on 
the same set of axes. What is the difference in phase between these? 

17. What is the difference in phase between the curves of y = sin x 
and y = cos x? Between y ■» cos x and y = sin (x + $*")? 

18. Plot the curve y — e~* sin x for values of x from to 2x. 

19. Plot the curve i « e~j* sin (2* -f- Jx) for values of t from —2 
to 8. 

20. In an oscillatory discharge of a condenser under certain con- 
ditions, the charge q at any time t is represented by the equation, 
q - 0.00224e-* 000 ' sin (8000* + tan" 1 2), where q is in coulombs and t 
in seconds. Plot the curve showing values of q for values of t from to 
0.0012 second. What is the period? 

Suggestion. — Choose 0.0001 second as a unit on the t-axis, and 0.001 
coulomb as a unit on the g-axis; and let the length representing a unit 
on the g-axis be about twice that for the unit on the t-axis. Plot the 
exponential curve first, and then the sine curve choosing as a unit on 
the g-axis the length representing 0.001 coulomb. 

EQUATIONS IN POLAR COORDINATES 

136. Discussion of the equation. — As in the case of equa- 
tions in rectangular coordinates, in polar coordinates the dis- 



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§137] OTHER LOCI AND EQUATIONS 175 

cussion of an equation helps greatly in learning the properties 
of the curve. The discussion is exactly similar to that in 
rectangular coordinates. 

(1) Intercepts. — (a) The intercepts on the polar axis are 
found by putting $ = 0°, 180°, 360°, • ••• nl80°. (b) The 
intercepts on the 90°-line are found by putting $ = 90°, 270°, 
etc. (c) Putting p =• and solving for 0, gives the values of $ 
for which the curve passes through the pole. 

(2) Symmetry. — (a) If the form of the equation does not 
change when — p is substituted for p, the curve is symmetrical 
with respect to the pole. (6) If it does not change when — $ 
is substituted for 0, the curve is symmetrical with respect to 
the polar axis, (c) If it does not change when ir — is 
substituted for 0, the curve is symmetrical with respect to 
the 90°-line. 

Show why each of these is true. Are their converses true? 
(See Art. 138.) 

(3) Extent. — If the equation is solved for p in terms of 0, the 
following can be determined: (a) Values of for which p has 
maximum or minimum values. In general this can be done 
readily when trigonometric functions are involved. (6) 
Values of for which p becomes infinite. These values 
determine the direction in which the curve extends to infinity, 
(c) Values of for which p is imaginary, that is, for which there 
is no curve. 

137* Loci of polar equations. — Since some of the conditions 
of the previous article are sufficient but not necessary, care 
must be taken in determining symmetry and extent of curves. 
On the whole, however, the plotting is very similar to that in 
rectangular coordinates, and is best illustrated by examples. 
It will be found convenient to use polar coordinate paper. 

Example 1. — Discuss and plot p = 1 -f- 2 sin 6. 

Discussion. — (1) Intercepts on polar axis, = 0, p = 1; 9 ■* 180°, 
p - 1. Intercepts on 90°-line, 6 - 90°, p - 3; $ = 270°, p = -1. 
When p = 0, sin $ - -i, and $ - 210° or 330°. 



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176 



ANALYTIC GEOMETRY 



[§137 



(2) Condition for symmetry with respect to the polar axis does not 
hold; but the curve is symmetrical with respect to the 90° -line since 
sin (x — 0) = sin 0. 

(3) Since p = 1 4- 2 sin 0, the maximum value of p will occur when 
sin = 1, or = 90°; and the minimum value of p will occur when sin 
$ = — 1, = 270°. No value of makes p imaginary. 

Plotting. — On account of the symmetry it is only necessary to find 
points for values of from 0° to 90° and from 270° to 360°. The curve 
is shown in Fig. 123. 




Fig. 123. 






sin 


P 


0° 


0.0 


1.00 


30° 


0.50 


2.00 


45° 


0,707 


2.41 


60° 


0.87 


2.73 


90° 


1.00 


3.00 


270° 


-1.00 


-1.00 


300° 


-0.87 


-0.73 


315° 


-0.707 


-0.41 


330° 


-0.50 


0.0 


345° 


-0.26 


0.48 


360° 


0.0 


1.00 



Example 2. — Discuss and plot p = a cos 20. The four-leafed rose. 

Discussion. — (1) Intercepts on polar axis, — 0, p - a; B = 180°, 
p = a. Intercepts on 90°-line, = 90°, p = -a; = 270°, p = —a. 
When p - 0, B - 45°, 135°, 225°, 315°. 

(2) Symmetrical with respect to the polar axis, and the 90°-line. 

(3) Since p = a cos 20, the maximum values of p occur when cos 20 = 1, 
or when 0.= 0° and 180°. The minimum values occur when = 90° 
and 270°. 

Plotting. — On account of symmetry find points for values of in 
the first quadrant. The curve is as shown in Fig. 124. The arrow 
heads indicate direction in which the curve is traced as increases from 
0° to 360°. 



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§138] 



OTHER LOCI AND EQUATIONS 



177 



e 


cos 20 


P 


0° 


1.0 


a 


10° 


0.94 


0.94a 


20° 


0.77 


0.77a 


30° 


0.50 


0.50a 


40° 


0.17 


0.17a 


45° 


0.0 


0.0 


60° 


-0.17 


-0.17a 


60° 


-0.50 


-0.50a 


70° 


-0.77 


-0.77a 


80° 


-0.94 


-0.94a 


90° 


-1.0 


—a 



Fig. 124. 

138. Remarks on loci of polar equations. — By definition, 
the locus of an equation requires that, (1) if the coordinates of 
any point satisfy the equation, the point is on the locus; 
(2) if any point is on the locus, the coordinates of this point 
satisfy the equation. 

In rectangular coordinates, there is no trouble in seeing that 
these conditions are fulfilled. This is because, in rectangular 
coordinates, there is one and only one point for every pair of 
coordinates; and, conversely, to every point there is just one 
pair of coordinates. 

In polar coordinates, trouble may arise since there is an 
ambiguity because a point has an indefinite number of pairs of 
coordinates determining it. 

Thus, in example 2 of the preceding article, it is seen that 
the point determined by the pairs of coordinates (\a } 60°), 
(-£a, 240°), (-Ja, -120°), and (Jo, -300°) is on the locus; 
but only (— Ja, 240°) and (— Ja, — 120°) satisfy the equation. 

For a like reason a curve may be symmetrical with respect 
to the polar axis even though (p, — 0) when substituted for 
(p, 0) changes the equation. In this case, some of the other 
pairs of coordinates of the point (p, — 0) would not change the 
equation. 

12 



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178 



ANALYTIC GEOMETRY 



[§139 



*»x 



139. Spirals. — Definition. — The locus of a point that 
revolves about a fixed point, and, at the same time, recedes 
from or approaches this point according to some law, is called 

a spiral. The fixed point is 
called the center of the spiral. 
When an angle is used in an 
equation and is not involved in 
a trigonometric function it is 
considered to be expressed in 
radians. 

Example.— Diacusa and plot the 
equation p = a # , p = a > 1. This is the 
logarithmic spiral. 

Discussion. — (1) When 0=0, p = l. 
(2) There is no symmetry. 
, p increases toward + » . As $ de- 
creases toward — », p approaches 0. 

Plotting. — The curve is readily plotted from a series of points. For 
a = 1.5 it is as shown in Fig. 125. 




Fig. 125. 
(3) As $ increases toward + < 



e 


-3 


-2 


-1 





1 


2 


3 


4 


5 


6 


p 


0.296 


0.444 


0.667 


1 


1.5 


2.25 


3.38 


5.06 


7.59 


11.39 



140. Polar equation of a locus. — The 

equation of a locus may often be f ound with 
greater ease in polar than in rectangular 
coordinates. The method is similar to 
that for finding the equation in rectangular 
coordinates, and has already been applied 
to the straight line and the conic sections. ° 
(See Arts. 69, 78, 91, 103, 117.) 

Example. — In Fig. 126, OT is the diameter of 
a fixed circle. At T a tangent is drawn, while 
about a secant revolves meeting the tangent in 
Q and the circle in R. The point P on the line 
OQ is taken so that OP = RQ. Find the equa- 
tion of the locus of P. 




Fig. 126. 



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§140] OTHER LOCI AND EQUATIONS 179 

Solution. — Choose' as pole, OT as polar axis, and let OT « 2a. Let 
the polar coordinates of P be (p, 0). 

Then since ORT is a right angle, PQ « OR - 2a cos 0. 

Since OTQ is a right angle, OQ =* — ~ 

COS 

tt a>» a>^ »•* 2a rt A 2a sin* 

Hence p - OP = OQ — PQ — 2a cos 0, or p « — . 

y ^ ^ cos ' COS 

This may be transformed to rectangular ' codrdinates and obtain 
I/ 1 — o » the equation of the cissoid of Diodes derived in article 126. 

&Qj ~~% 

Compare the derivations of the equation by the two methods. 

EXERCISES ; 
Discuss and plot the following equations: 
*• p s f" Z a' ^ parabola. 

2. p sin tan $ ^ 4a. A parabola. 

3. p* cos 20 = a*. An equilateral hyperbola. 

4. p = 3 cos + 2. Transform to rectangular codrdinates. 
6. p = a tan* sec 0. Semi-cubical parabola. 

6. p — a cot* esc 0, Semi-cubical parabola. 

7. Transform equations of exercises 5 and 6 to rectangular codrdinates 
and compare with article 123. 

8. p = a — 6 sin when a<b, when a = 6, and when a>b. Limacons 
of Pascal. 

9. p* = a* cos 30. Is the curve symmetrical with respect to the 
90°-line? Does the test apply? 

Sketch the following roses by first drawing the radial lines corre- 
sponding to values of which make p «= 0, and for values of which 
make p maximum in numerical value; and then determining the changes 
in the values of p between these successive values of 0. 

10. p — a sin 20. 11. p — a sin 30. 12. p = a sin 40. 
13. p = a cos 30. 14. p = a cos 40. 15. p — a cos 50. 

In plotting the curves of the following equations, it should be noted 
that in polar codrdinates it is sometimes necessary to carry the angle 
beyond 360° in order to secure the complete locus. 

16. p — a sin 3 ^0. 17. p — a sin J0. 18. p* cos «= a* sin 30. 

19. p = a(sin 20 -f cos 20). 20. p* — a* sin J0. 

21. p = a(l ± cos 0). The cardioids. 

Discuss and plot the following spirals: 
. 22. p0 = a. Hyperbolic or reciprocal spiral. 
L ' 23. p = aO. Spiral of Archimedes. 

24. p s =* a0. Parabolic spiral. 



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180 ANALYTIC GEOMETRY [§141 

25. />*$ =* a. The lituus or trumpet. * 

26. Derive the equation of the locus of a point such that: 

(1) Its radius vector is inversely proportional to its vectorial angle. 

Ans. The hyperbolic spiral. 

(2) Its radius vector is directly proportional to its vectorial angle. 

Ans. The spiral of Archimedes. 

(3) The square of its radius vector is directly proportional to its 
vectorial angle. Ans. The parabolic spiral. 

(4) The square of its radius vector is inversely proportional to its 
vectorial angle. Ans. The lituus. 

(5) The logarithm of its radius vector is directly proportional to its 
vectorial angle. Ans. The logarithmic spiral. 

27. Find the equation of the locus of the midpoints of the chords of 
the circle p = 2r cos 0, and passing through the pole. 

28. Chords of the circle p = 2r cos $ and passing through the pole 
are extended a distance 2b. Find the equation of the locus of the 
extremities. 

PARAMETRIC EQUATIONS OF LOCI 

141. Parametric equations. — When the coordinates of 
points on a locus are expressed separately as functions of a 
third variable, these equations are called the parametric 
equations of the locus. 

The new variable introduced in finding the parametric 
equations is called a parameter. 

The parameter may be introduced either for convenience or 
as a necessity, since in some cases it is easier to obtain the 
coordinates of points on a locus as functions of a third variable 
than it is to obtain a single equation connecting the coordi- 
nates of the points; and frequently two equations using the 
parameter can be obtained where it is not possible to obtain 
a single equation connecting the two variables. 

As will be seen, the parameter can be chosen in a great 
variety of ways, but it is usually chosen because of some 
simple geometric relation, or it is the time during which the 
point tracing the curve has been in motion. 

Example 1. — The parametric equations x = x\ + nt and y =» y x + mi 
represent the straight line which passes through the point (xi, y{) and 

has the slope — . 
n 



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\ 



^ ; <-* 



X 



^ 



- -f 



§141] 



OTHER LOCI AND EQUATIONS 



181 



That this is so can be seen by assigning values to t and plotting the 
values of x and y, or by eliminating t and obtaining the equation 



y 



y x - — (* - xi) f 



£(+*) 




*-x 



Fig. 127. 



which is the equation of a straight line. 

Example 2. — Consider a circle with center 
at the origin and radius r as generated by a 
point P starting on the x-axis and moving 
counter-clockwise. Then it is evident from 
Fig. 127 and the definitions of the sine and 
cosine, that the parametric equations 

x = r cos $ and y « r sin 0, ^ 

where is the angle generated by the radius to the point P, represent the 
circle. 

Also, squaring and adding the equations, X s + y* ■■ r*. 

Example 3. — The equations x = t* and y = 2t are parametric equa- 
tions of the parabola y* — 4a;, as can be seen by eliminating t from the 
two equations. The curve can be plotted by assigning values to t and 
computing the corresponding values of x and y. 



t 


- 4 


-3 


-2 


-1 


-i 





i 


1 


2 


3 


4 


x 


16 


9 


4 


1 


i 





i 


1 


4 


9 


16 


y 


- 8 


-6 


-4 


-2 


-1 





1 


2 


4 


6 


8 




Fig. 128. 



The values of x and y are plotted, and 
the curve is as shown in Fig. 128. It is 
observed that as t varies from — » to 
+ oo the corresponding point will trace 
out the curve, coming from » on the 
lower half and going to <» on the upper 
half of the parabola. 

Example 4. — The equations x=*a cos $ 
and y = b sin $ are parametric equa- 
tions of the ellipse as is shown in 
article 104. 

Example 5. — The equations x = a sec $ 
and y = b tan are parametric equa- 
tions of the hyperbola; for dividing 



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182 ANALYTIC GEOMETRY [§142 

the first by a, the second by 6, squaring, and subtracting the second 
from the first gives 

JC* y* 

~i — fi = sec* — tan* 0*1. 

Example 6. — Equations (1) given in article 94, x = v cos at and 
y = v sin at — igt* t are parametric equations of a parabola. Here 
t is the number of seconds the point has been moving. 

EXERCISES 

1. Write parametric equations of the straight line through (—3, 2) 
and having a slope of 2. f lot the line from these equations. 

2. Write parametric equations of the circle with center at (2, 3) and 
radius 5. 

3. Represent the parabola y* = 4a; by several pairs of parametric 
equations. 

Suggestion. — Either x or y can be represented at pleasure, but the 
other must be dete rmined in accordance with this. For instance, if 
x - t* + 1, y - 2 V** + 1. 

Plot the following parametric equations. In each case eliminate 
the parameter and find a single equation representing the same curve. 

4. x = 4 - t\ y - t - 1. 

5. x = 5 cos 0, y — 3 sin 0. 

6. s — 2 4* sin 0, y = 2 cos 0. 

7. s = e + %\ y = e - P. 

8. 3 = 5+2 cos 0, y=4+3sin0. 

9. x = 1 — cos 0, y = J sin J0. 
_10. s = cos 0, y = cos 20. 

11. x = a sin -f & cos 0, y = a cos — b sin 0. , 

12. x = a cos 1 0, y = 6 sin 8 0. 

142. The cycloid. — The plane curve traced by a fixed point 
on a circle as the circle rolls along a fixed straight line is 
called a cycloid. The rolling circle is called the generator 
circle and the fixed straight line the base. 

The parametric equations of the cycloid can be derived as 
follows: 

In Fig. 129, let OX be the fixed straight line, C the generator 
circle of radius a, and P{x, y) the tracing point. Also suppose 
the circle is rolling towards the right. - 

Choose OX as the x-axis and the origin where the tracing 



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§143] 



OTHER LOCI AND EQUATIONS 



183 



point is in contact with the fixed line. Also choose as para- 
meter the angle 0, through which the radius to the tracing 
point turns. Draw the lines shown in the figure. 

Then x = ON - OM - NM = OM - PQ, 
and y = NP = MC- QC. 




[40] 



Fig. 129. 

But OM = arc MP = a$ f PQ = a sin 6, MC « a, and 
QC = a cos 0. 

Substituting these values gives 

x = a(8 — sin 8), 
y = a(l — cos 6). 
These are the forms of the equations most frequently used 
in dealing with the cycloid. If is eliminated the equation 
in x and y is 

x = a vers -1 - — y/2ay — y*, 

a form that is seldom used. 

EXERCISES 

1. Plot the cycloid from the parametric equations. Is the curve 
periodic? 

2. Construct a figure in which 9O°<0<18O°, and derive the equation 
of the cycloid from it. 

3. Derive parametric equations for the locus traced by a point on a 
fixed radius and at a distance b from the center of the circle rolling as 
in generating the cycloid. First, suppose b<a; second, suppose b>a. 

4. Plot the curves of exercise 3. Such curves are called trochoids. 

143. The hypocycloid. — The plane curve traced by a fixed 
point on a circle as the circle rolls along a fixed circle internally 
is called an hypocycloid. 



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184 



ANALYTIC GEOMETRY 



[§U3 



The derivation of the parametric equations is as follows: 
In Fig. 130, let be the fixed circle with radius a, and C the 

generator circle with radius 6. 
Let P(x, y) be the tracing 
point. 

Choose as origin and OX 
as x-axis. Also let the trac- 
ing point start at A where 
the x-axis intersects the fixed 
• circle. Choose as parameters 
the angle 6, through which 
the line of centers of the two 
circles turns, and the angle <p, 
through which the radius of 
the generator circle turns. 




Fig. 130. 



Draw the lines shown in the figure. 

Then x = OM - OB + BM = OB.+ NP, 
and y = MP = BN - BC - NC. 

But OB = OC cos 6 = (a - b) cos 6, 
and NP = CP sin PCN = b cos (<p - 6). 

Also BC = OC sin 6 - (a - 6) sin 0, 
and JVC = CP cos PCN = 6 sin (*> - 0). 

Substituting these values gives 

x = (a — b) cos + 6 cos (<p — 0), 
and y = (a — 6) sin — 6 sin ($> — 0). 

To eliminate the parameter <p } notice that 

arc AQ = arc PQ, or a0 = 6??, and hence ^ = 



ad 



Substituting the value of <p in the above equations, 

/a-b\ 



t«] 



x = (a — b) cos 8 + b cos ( — =- — j e, 
y = (a — b) sin 8 — b sin ( a 7" ) 6. 



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§144] 



OTHER LOCI AND EQUATIONS 



185 




Fio. 131. 



The hypocycloid is a closed curve only when the diameters 
of the two circles are commensurable. 

If o = 26, equations [41] become x = a cos 6 and y = 0. 

Therefore when the radius of the generator circle is one- 
half the radius of the fixed circle, the tracing point moves in 
a straight line. 

The most important special case of the hypocycloid is the 
four-cusped hypocycloid, in which a = 46. The curve is 
shown in Fig. 131. Here the parameter 
can be eliminated and a single equation 
in x and y obtained. 

Putting 6 = Jo in equations [41], 

x = fa cos0 + 1<* cos 30, 

and y = \a sin0 — \a sin 30. 

But from trigonometry 

cos 36 = 4 cos 8 6 — 3 cos 6, 
and sin 36 = 3 sin 6 — 4 sin 8 0. 

Substituting and simplifying, x = a cos 8 6 and y = a sin* 6. 
Affecting by the exponent | and adding, gives the equation 
in x and y, 

[42] x 1 + y 1 = a 1 . 

144. The epicycloid. — The plane curve traced by a fixed 
point on a circle as the circle rolls along a fixed circle externally 
is called an epicycloid. 

Using a and 6 as the radii of the fixed circle and the generator 
circle respectively, and 6 and <p as shown in Fig. 132, the 
equations of the epicycloid are 

x = (a + b) cos e - b cos(?-^W 
[43] 

y - (a + b) sin 8 - b sin (— £^) *• 

An important special case of the epicycloid is the cardioid, in 



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186 



ANALYTIC GEOMETRY 



[§145 



which a = b. The curve is as shown in Fig. 133. The 
equations here become 

x — 2a cos 6 — a cos 20, 
and y = 2a sin — a sin 20. 



P(*.V) 



r* 





Fig. 132. 



Fig. 133. 



P(*.v) 



+X 



145. The involute of a circle. — If a string is wound around 
a circle, the curve in the plane of the circle^ traced by a point 

on the string as it is unwound 
and kept taut, is called the 
involute of the circle. 

The parametric equations 
may be derived as follows: 

Choose the x-axis through the 
point where the tracing point is 
in contact with the circle, and 
the origin at its center. The 
parameter 0, Fig. 134, is the 
angle through which the radius 
to the point of tangency of the string has turned. 
Then from the figure, 

x = OM = OB + BM = OB+LP, 
and y = MP = BL = BT - LT 

But OB = a cos 0, LP = TP sin 6 = aO sin $, 
and BT = a sin 0, LT = TP cos 8 = aO cos 6. 




Fig. 134. 



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§145] OTHER LOCI AND EQUATIONS 187 

Substituting these values gives 
r^-, x = a cos 8 + ad sin 6 

y = a sin 6 — ad cos 6. 

EXERCISES 

1. Derive the equation of the f our-cusped hypocycloid. 

2. Derive the equation of the epicycloid. 

3. Derive the polar form of the equation of the cardioid from the 
parametric equations given in article 144. 

Suggestion. — In the polar form of the equation the pole is at A, Fig. 
133. Notice that Z XAP .— Z XOC, and hence the parameter $ is 
equal to the polar coordinate 0. 

First, square and add the equations of Article 144, then translate to 
new origin at A (a, 0), finally, transform to polar coordinates and de- 
rive the equation p = 2a (1 — cos $). 



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CHAPTER X 
EMPIRICAL LOCI AND EQUATIONS 

146. General statement. — In common every day affairs, in 
business, in the sciences as physics, chemistry, and biology, 
and in engineering, questions often arise involving the relations 
of variables. Values of these variables can be plotted accord- 
ing to some system of codrdinates, and, in this manner, curves 
obtained that give valuable information. Often the desired 
facts can be discovered directly from the curve; but frequently, 
especially in the sciences and in engineering, it is of the utmost 
importance to find a mathematical equation representing the 
curve more or less accurately. 

The determination of the equation may be a comparatively 
simple matter, but often it is very laborious and involves 
methods beyond the scope of this text. 

A curve that is plotted from observed values of the related 
variables is called an empirical curve or locus. 

The equation of an empirical curve is an empirical equation. 

Usually the empirical equation represents a curve that only 
approximates the empirical curve more or less accurately. 

147. Empirical curves. — Innumerable examples of empiri- 
cal curves could be given. For many of these there may be no 
necessity nor reason for finding equations representing them. 

The rise and fall in the price of a certain stock may be 
represented graphically by using the price each day as the 
ordinate of a point of which the date is the abscissa. ( The 
curve drawn through these points win show at a glance the 
fluctuations of this particular stock. 

If the weight of a child is taken from month to month, a 
curve can be plotted by using the weights a3 ordinates and 
the corresponding dates as abscissas of points. 

188 



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§147] EMPIRICAL LOCI AND EQUATIONS 189 

Empirical curves are often traced mechanically by instru- 
ments designed for that particular purpose. In this manner, 
at a weather bureau station, a curve is traced showing the 
relation between the temperature and the time. In Fig. 135, 
is a similar curve that shows the per cent of carbon dioxide 



Fig. 135. 

in the flue gas from a power plant. The variables are the 
time and the per cent of carbon dioxide. The system of 
coordinates is apparent. 

In Fig. 136, are plotted several curves showing the changes 
in the cost of living from July, 1914 to November, 1919. The 
data was taken from the Research Report issued by the 
National Industrial Conference Board. 



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190 



ANALYTIC GEOMETRY 



[§148 



In such curves as these the information desired is gained 
directly from the curve, and no attempt would be made to 
derive an equation. 




1914 



1915 Year 1916 



1917 



1918 



1919 



Change in prices from 1014 to 1019. 
(1) Shelter; (2) Heat and light; (3) Sundries; (4) Cost of living; (5) Food; (6) Clothing. 

Fig. 13ft. 

148. Experimental data. — In laboratory experiments and 
practical tests, pairs of simultaneous values of two varying 
quantities are measured. When these pairs of values are 
plotted, a curve is determined from which useful information 
may be obtained. The problem of finding the empirical 
equations representing such curves will now be considered. 

All data that is a result of measurements must be assumed 
to be subject to some degree of error, hence the endeavor will 
always be to approximate as closely as possible, both in the 
curve and in the equation. 

Sometimes in a problem of this kind the general form of 
the equation of the curve is known beforehand, and sometimes 
nothing at all is known but the coordinates measured in the 



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§149] EMPIRICAL LOCI AND EQUATIONS 191 

experiment. If the general form of the equation is known, the 
computations for finding the definite equation can be made at 
once; but, if the general form of the equation is not known, the 
points are plotted so as to discover the general form if possible. 

149. General forms of equations. — The f orms of equations 
frequently used are the following. Most of these have been 
studied in previous chapters, and should be reviewed, if 
necessary, so that their forms may be clearly in mind. 

(1) y = mx + by straight line. 

(2) y = cx n , n>0, parabolic type. 

(3) y = cx n , n<0, hyperbolic type. 

(4) y = ab x or y = ae* x , exponential type. 

(5) y = a + bx + ex 2 + dx* + • • • + qxP. 

For the parabolic type it is often necessary to use 

y — k = c(x — h) n , n > 0, 
where the vertex is at the point (A, h) ; and for the hyperbolic 
type 

y — k = c(x — h) n , n < 0. 

150. Straight line, y = mx + b. — This is the form of the 
empirical equation when it is known that the relation between 
the variables is that of a direct variation. Since in the 
equation y = mx + b there are but two arbitrary constants, 
two pairs of measured values would be sufficient to deter- 
mine the equation completely provided the values could be 
measured accurately. Since this is not possible, a larger 
number of pairs of values are measured, and from these an 
equation is determined that represents the straight line lying 
most nearly to all the points. The method used in the follow- 
ing example for securing the equation ij called the method of 
least squares. The theory underlying the method is too 
difficult to be given here. 

Example. — Find the equation of the straight line that is in the form 
y — mx -f- 6, lying most nearly to the points determined by the follow- 
ing measured values of x and y: 



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192 



ANALYTIC GEOMETRY 



[§150 



X 


40 


50 


62.4 


70 


80.5 


90 


97 


V 


8.7 


7.5 


6.5 


5.85 


5.05 


4.25 


3.75 



Solution. — Here the type of the equation is given so there is no need 
of plotting the points. 

Substituting each pair of values successively in the equation 
y — mx +b gives the seven equations: 

8.7 = 40m +6, 
7.5 - 50m + 6, 
6.5 - 62.4m + 6, 
5.85= 70m + 6, 
5.05 - 80.5m + 6, 
4.25= 90m + 6, 
3.75= 97m + 6. 
Multiplying each of these by the coefficient of m in that equation 
and adding the seven resulting equations, gives 

2690.875 = 36883.01m + 489.96. (1) 

r 

15 



10 



10 20 



80 40 60 60 
Fig. 137. 



70 80 90 100 



Multiplying each of the seven equations by the coefficient of b in 
that equation and adding the results, gives 

41.6 = 489.9m + 76. (2) 

Solving (1) and (2) for m and 6, gives 

m = -0.085, and b = 11.89. 
Substituting these values in y = mx «+■ 6, gives 

y = -0.085* + 11.89. (3) 

This is taken as the equation of the straight line lying most nearly 
to all the points. 

In Fig. 137 are plotted equation (3) and the points whose 
coordinates are the observed values of x and y. 



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§151] 



EMPIRICAL LOCI AND EQUATIONS 



193 



151. Method of least squares. — The method of least 
squares given in the previous article becomes tedious when 
there are many observations and the numbers are large. A 
sufficiently accurate result may be obtained by plotting the 
points, and obtaining the arbitrary constants of the equation 
by using two points that lie on the straight line that appears 
to be the best. If none of the plotted points lie on this line, 
use coordinates of points that do lie on the line. 

The method of least squares is quite mechanical, while the 
best straight line if determined by plotting is a matter of 
judgment and a good eye. 

The method by least squares for finding the empirical 
equation is stated in the following: 

Rule. — First, substitute each pair of observed values of the 
variables in the general equation. 

Second, if there are just as many equations as there are con- 
stants to be found, solve these equations for the constants. If 
there are more equations than there are constants, multiply each 
equation by the coefficient of the first constant in that equation, 
and add the resulting equations to form one equation. Proceed 
likewise for each other constant, and thus find as many equations 
as there are constants. 

Third, solve these equations for the constants. 

Fourth, substitute the constants, thus found, in the general 
equation and obtain the required empirical equation. 

EXERCISES 

1. A wire under tension is found by experiment to stretch an amount I, 
in inches, under a tension T t in pounds, as given in the following table. 
Assume the relation I — kT (Hooke's law) and find the equation which 
best represents the relation between I and T. 



T 


5 


10 


20 


30 


40 


50 


I 


0.003 


0.009 


0.019 


0.030 


0.040 


0.05S 



13 



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194 



ANALYTIC GEOMETRY 



[§151 



2. Find the empirical equation in the form of y « mx + b best 
representing the relations between the values given in the following 
table: 



X 


12 


15.3 


17.8 


19 


y 


24.4 


29 


32.6 


34.2 



3. Find the equation of the straight line lying most nearly to the 
points determined by the following pairs of measured values: 



X 


12 


15 


18 


21 


24 


y 


24.4 


28.6 


32.7 


37.1 


41.2 



4. Find the -empirical equation connecting R and t from the following 
table of experimental values. R is in ohms and t in degrees centigrade. 
The equation is assumed to be in the form R = mt *+ b. 



t 


10.1 


15 


21 


26.8 


33.1 


40.4 


R 


9.907 


9.923 


9.940 


9.959 


9.979 


10.002 



5. Find the empirical equation giving H in terms of t, from the data 
of the following table. H is the total heat in a pound of saturated 
steam at t degrees centigrade. The general form is H = mt + b. 



t 


65 


85 


100 


110 


120 


H 


626.3 


632.4 


637 


640.9 


643.1 



6. In an experiment with a Weston differential pulley block, the 
effort Ej in pounds, required to lift a weight W, in pounds, was found 
to be as follows: 



w 


10 


20 


30 


40 


50 


70 


90 


100 


E 


3.25 


4.875 


6.25 


7.5 


9 


12.25 


15 


16.5 



Find the empirical equation in the form E = mW + 6. 



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§152] 



EMPIRICAL LOCI AND EQUATIONS 



195 



7. Plot the data given in exercise 6, and draw a line that, in your 
judgment, lies most nearly to all the points. Select two points that 
lie as nearly on the line as any, and determine the equation from the 
coordinates of these points. Compare the result with that of exercise 6. 

8. In the following table, W is the weight of potassium bromide which 
will dissolve in 100 grams of water at t degrees centigrade. Find the 
empirical formula in the form W = mt + 6, connecting W and t 



t 





20 


40 


60 


80 


w 


53.4 


64.6 


74.6 


84.'7 


93.5 



152. Parabolic type, y = ex 11 , n > 0. — If it is not known 
that the general form of the equation is of some particular 
type, it is well to plot the data on rectangular coordinate paper 
and judge the type from the curve. After the general form is 
selected, it is often difficult to determine whether or not it 
actually represents the observed values with sufficient accuracy 
for the purposes of the problem. A device that is of great 
assistance in determining whether to retain or reject the type 
selected is to transform the general equation into a linear 
equation, and see if the data plots as a straight line* This is 
done as follows when the equation is of the parabolic type: 

Given equation, y = cxP. 

Taking logarithms of both sides, 

log y = log c + n log x, 

which is a linear equation in log x and log y. If the points 
with coordinates (log x y log y), where x and y for each point 
are a pair of observed values, are plotted, and these points 
are found to lie approximately on a straight line, then the 
general form of the equation is suitable to the problem. 

The values of the constants, log c and n, can be determined 
by the method of least squares as in article 151. 

Of course, if the general form of the equation to be used is 
known to be of the parabolic type, the plotting is not necessary. 



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196 



ANALYTIC GEOMETRY 



[§152 



Example.— Q is the quantity of water, in cubic feet per second, that 
flows through a right isosceles triangular notch when the surface of the 
still water is at a height H feet above the bottom of the notch. The 
values of H and Q in the following table are measured. Find the equa- 
tion connecting H and Q. 



H 


l 


1.5 


2 


2.5 


3 


4 


Q 


2.63 


7.25 


15 


26 


41 


84.4 



Solution. — The values of H and Q are plotted in Fig. 138, and con- 
nected with the curve, which appears to be of the parabolic type. 

Assume the general form Q = cH n . (1) 

Taking logarithms, log Q = log c + n log H. (2) 

Put log H = x, log Q = y and log c = 6, and the equation becomes 

y » nx + b. 

Q Y 




a.6 

.2 

1.5 
.1 































































Fig. 138. 



.2 .4 .6 .i 
Fig. 139. 



The values of x and y are found and plotted in Fig. 139. The points 
lie approximately in a straight line. 



logil 





0.1761 


0.3010 


0.3979 


0.4771 


0.6021 


log<? 


0.4200 


0.8603 


1.1761 


1.4150 


1.6128 


1.9263 



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§163] EMPIRICAL LOCI AND EQUATIONS 197 

Substituting the values for x and y in the equation y — nx + b, the 
following equations are obtained: 

0.4200 - On +6, 

0.8603 »0.1761n + fc, 
1.1761 =0.3010n +6, 
1.4150 = 0.3979n + 6, 
1.6128 = 0.4771n + 6, 
1.9263 =0.6021n +6. 

Solving these equations by the method of least squares, 

n - 2.5 and b - 0.4208. 

Substituting in equation (2), log Q = 0.4208 + 2.5 log H. 

Then log Q =* log 2.635 + log H*K 

Or log Q - log (2.635ff 1 *). 

.". Q — 2.635ff*- 6 , the required equation. 

The equation can be tested by computing values of Q for the several 
observed values of H f and comparing with the observed values of Q. 

153. Hyperbolic type y = cx n , n < 0. — Data that are 
known to give an equation of this type can be handled in 
precisely the same manner as the parabolic type. The only 
difference that will arise Will be that the value of n is 
negative. 

154. Exponential type, y = ab x or y =* ae 1 ". — The data 
from certain experiments, such as those involving friction, 
give rise to exponential equations. As with the other types 
the data can be plotted on rectangular coordinate paper and 
the general form of the equation determined. If it is thought 
to be of the exponential type, it can be tested by taking the 
logarithms of both sides of the equation and plotting on 
rectangular coordinate paper. If the points lie on a straight 
line, the assumed equation is correct. 

In order to express the form y = ab s in the form y = ae** it is 
only necessary to put 6 = 6*, whence log b = k log e, or 

k = j^Si = 2.3026 log b. 
loge 



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198 



ANALYTIC GEOMETRY 



[§154 



Example. — From the following data determine the relation between 
W and 0. 



e 


1.57 


3.14 


4.71 


6.28 


7.85 


9.42 


11 


W 


5.35 


7.15 


9.55 


12.8 


17.12 


22.9 


30.8 



Solution. — First, plot the data given and determine the form of the 
equation to be used. The plotting is shown in Fig. 140, and the equation 
assumed is 

W - ab . (1) 

Second, to test this, take the logarithms of both sides of W = ab e . 

This gives log W = log a + B log b. 

Put log W = y, log a = B, and log b = m. 

This gives y = m$ + B. (2) 



W 



30 



20 



10 



















—e 



O 5 10 

Fig. 140. 











1.5 




/ 


/ 


1 


/ 






.5 



s* 










9 



O 5 

Fig. 141. 



Arranging these values in a table and plotting, gives approximately 
a straight line as shown in Fig. 141. 



e 


1.57 


3.14 


4.71 


6.28 


7.85 


9.42 


11 


y 


0.728 


0.854 


0.980 


1.107 


1.234 


1.360 


1.489 



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§165] 



EMPIRICAL LOCI AND EQUATIONS 



199 



Substituting the pairs of values of $ and y in equation (2), 
0.728 - 1.57m + B, 
0.854 = 3.14m + B, 
0.980 - 4.71m + B, 
1.107 « 6.28m + B, 
1.234 - 7.85m + B, 
1.360 = 9.42m + B, 
1.489= 11m + B. 

Solving these by the method of least squares, gives 
m = 0.0807 and B = 0.6005. 
Then a = 3.985 and b = 1.204. 

.'. W = 3.985 X 1.204 9 , the required equation. 

This expressed in the form W — ae** gives, 
W - 3.985e«. lw **. 

155. Probability Curve, y = ae~ te *. — The curve that is 
perhaps the most widely used of any in dealing with experi- 
mental data is one variously called "the probability curve," 
"the error curve," and "the normal distribution curve." 
It is represented by the equation 

y = ae^ x \ 

where a and b are constants to be determined from the data. 
It is evidently symmetrical with respect to the y-axis. While 




definite uses of this curve are beyond the scope of this chapter, 
it may be stated that it is used wherever a most probable 
correct value is to be determined from a large number of 
independent measurements or observations. It is used in 



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200 



ANALYTIC GEOMETRY 



[§166 



the study of statistics, in astronomy, biology, and chemistry, 
and in the study of theory of measurements. 
The form of the curve is shown in Fig. 142. 
166. Logarithmic paper. — Because of the frequent occur- 
rence of formulas of the parabolic and hyperbolic types, con- 
siderable use is made in 
engineering practice of 
logarithmic paper, that is, 
paper that is ruled in lines 
whose distances, horizon- 
tally and vertically, are 
proportional to the loga- 
rithms of the numbers 1, 
2, 3, etc. 

Logarithmic paper can 
be used instead of actually 
looking up the logarithms 
of the numbers as was 
done in the example of 
article 152. For if the 
values of H and Q are 
plotted as shown in Fig. 
143, a straight line is de- 
termined just as when the 
logarithms of H and Q 
were plotted on rectangu- 
lar coordinate paper. 

Semi-logarithmic paper 
is ruled uniformly the 
same as ordinary coordi- 
nate paper in one direction, 
and in lines spaced as on logarithmic paper in the other direc- 
tion. Semi-logarithmic paper may be used to advantage when 
testing an exponential type. In Fig. 144, the values of 6 and 
W of the example of article 154 are plotted into a straight line. 



in 




















90 






































80 






































10 






































60 






































60 






































40 






































30 
26 

SO 
16 

10 
















































































































9 








































8 






































7 






































6 


























































6 




















4 

8,5 

3 
















^~ ■ 


























































2,5 
2 

1.5 
1 

























































1,5 2 



2,5 3 4 5 6 7 8 9 
Fig. 143. 



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§156] 



EMPIRICAL LOCI AND EQUATIONS 



201 



EXERCISES 

1. Solve the equations of article 152 by the method of least squares 
and check the results given. 

2. Determine the equation of the hyperbolic type connecting x and 
y from the following pairs of values: 



X 


1.5 


2.8 


5.6 


8.3 


y 


0.573 


0.243 


0.094 


0.055 


m w 




18 






















































70 
60 
60 
40 

80 
































































































































^ 






























20 
10 




























9 










^9 


















8 








r-* 


** 


















7 
6 
5 








n^ 




























• 












































4 
3 






















































2 




























1 








i 


















9 



2 8 4 6 6 7 
Fio. 144. 



10 11 12 



3. In propelling a ship of a certain class at 10 knots, the following 
pairs of values of D and H are measured, where D is the displacement 
in tons and H is the indicated horse-power. Find a formula of the 
parabolic type connecting D and H. 



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202 



ANALYTIC GEOMETRY 



t§156 



D 


1100 


1530 


1820 


2500 


3130 


H 


440 


550 


620 


770 


890 



Compute H when D = 2000. 

4. For different heights, h in feet above the surface of the earth, 
the reading, p in inches, of the barometer are taken as given in the 
following table. Determine a formula of the form p — cut* connecting 
p and h. 



h 


' 


886 


2753 


4763 


6942 


V 


30 


29 


27 


25 


23 



5. The data of the example of article 154 was taken in an experiment 
to determine the coefficient of friction /i, when a cord is wrapped around 
a cylindrical shaft, Fig. 145. In performing the 
experiment, the cord has a weight of 2 pounds 
attached to one end, and a pull of W pounds at 
the other end induces slipping when the arc of 
contact is radians. Determine the value of p 
for the equation W — aeP°. /i = & of Art. 164. 

6. In testing the lubrication of certain oils in 
a bearing, 4} inches in diameter and 8 inches 
long with 250 revolutions per minute, the follow- 
ing pairs of values were measured, where p is the pressure in pounds 
per square inch and /i is the coefficient of friction. Determine a formula 
of the fprm n = ap n connecting p and /i* 




1 

p ; 


65 


115 


215 


315 


465 


'i 


0.0090 


;0.0056 


0,0036 


.0.0028 


0.0025 


Plot the values showing that the curve is of the hyperbolic type. 
7. In the same experiment as in exercise 6, but using another oil, the 
following values were obtained. Determine a formula connecting p and /*• 


V 


65 


115 


215 


315 


415 


515 


M 


0.00788 


0.00528 


0.00338 


0.00267 


0.00235 


0.00215 



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§157] 



EMPIRICAL LOCI AND EQUATIONS 



203 



8. In the following table are given the measurements taken in an 
experiment on friction, where /i is the coefficient of friction in a certain 
bearing running at a velocity of V feet per minute. Determine a formula 
of the form n — dV n connecting V and /i. 



V 


105 


157 


209 


262 


314 


366 


419 


471 


M 


0.0018 


0.0021 


0.0025 


0.0028 


0.003 


0.0033 


0.0036 


0.004 


9. In a mixture in a cylinder of a gas-engine, under adiabatic expan- 
sion, the following pairs of values are measured. Determine a formula 
in the form pv n = C connecting v and p. 


V 


0.8 


2 


4 


6 


9 


V 


200 


57 


22 


12.6 


7.2 



157. Empirical formulas of the type y = a + bx + ex 2 + 
dx 3 + • • • + qx n . — When a given set of corresponding 
pairs of values will not satisfy, in a satisfactory manner, any 
of the type equations already considered, the general equation 

y = a + bx + ex 2 + dx* + • • • + q& 

may be assumed. By substituting pairs of values in this 
equation, enough equations can be obtained to determine the 
constants a, 6, c, • • • . 

Since there must be, at least, as many equations as con- 
stants, no more terms can be assumed than the number of 
pairs of values measured. If there are more pairs of values 
than the number of terms assumed, the equations can be 
solved by the method of least squares. A less accurate 
method, but one more easily carried out, is to select as many 
of the equations as there are constants, and solve these for 
the constants. The equation thus found can be tested by 
substituting the pairs of values not used in the equations 
that are solved for the constants. 

If the points when plotted suggest a parabola, only three 



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204 



ANALYTIC GEOMETRY 



t§157 



terms need be used. If the arrangement of the points is 
more irregular, more terms must be assumed. 

Example. — The following measurements at different depths were made 
to determine the rate of flow in a river, where x is the fractional part 
of the depth from the surface and y is the rate of flow. Determine 
a formula of the form y ** a + bx + ex 1 connecting x and y. 



X 





0.2 


0.3 


0.4 


0.6 


0.8 


0.9 


y 


3.195 


3.253 


3.261 


3.252 


3.181 


3.059 


2.976 



Solution. — Substituting the pairs of values in y — a + bx + ex*, 

3.195 -a + 06 + 0c, 
3.253 = a +0.26 + 0.04c, 
3.261 = a +0.36 + 0.09c, 
3.252 -a +0.46 +0. 16c, 
3.181 - a +0.66 + 0.36c, 
3.059 - a +0.86 + 0.64c, 
2.976 = a + 0.96 + 0.81c. 

These are solved by the method of least squares as follows: 
Multiplying each by its coefficient of a and adding the seven resulting 
equations, gives 

22.177 = la + 3.26 + 2.1c. (1) 

Multiplying each by its coefficient of 6 and adding, gives 

9.9639 - 3.2a + 2.16 + 1.556c. (2) 

Multiplying each by its coefficient of c and adding, gives 

6.45741 = 2.1a + 1.5566 + 1.2306c. (3) 

Solving equations (1), (2), and (3) for a, 6, and c, 

a - 3.196, 6 - 0.438, c - -0.7608. 
Substituting these values in y = a + bx + ex 1 , gives 
y - 3.196 -«- 0.438* - 0.7608a; 1 , 
which is the required equation. 

EXERCISES 

1. Plot the corresponding pairs of values of x and y given in the 
example of article 157, and draw a smooth curve lying as near as possible 



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§157] 



EMPIRICAL LOCI AND EQUATIONS 



205 



to all the points. Select the three points lying most nearly on the curve, 
and use the coordinates of these to find the values of a, b, and c. Com- 
pare with the result given in the solution by the method of least squares. 
2. Determine an equation of the form y - a + bx + ex* + dx* from 
the following experimental values. Solve both by the method of least 
squares and by using the coordinates of four points. 



X 


0.4 


0.6 


0.8 


1.0 


1.2 


1.4 


1.6 


V 


0.89 


1.35 


1.96 


2.72 


3.62 


4.63 


5.76 



3. The melting point of an alloy of lead and tin containing x per cent 
of lead is t degrees centigrade. From the following table of measured 
values, find a formula in the form t = a + bx + ex*, giving the melting 
point of an alloy containing any known per cent of lead from 90 per 
cent to 35 per cent. 



X 


87.5 


84 


77.8 


63.7 


46.7 


36.9 


t 


292 


283 


270 


235 


197 


181 



For a further discussion of the subject of this chapter, the following 
works may be consulted: Merriman, Method of Least Squares; Weld, 
Theory of Errors and Least Squares; Johnson, Theory of Errors and 
Method of Least Squares; Palmer, Theory of Measurements; Steinmetz, 
Engineering Mathematics; Running, Empirical Formulas; Lipka, Cfraphr 
teal and Mechanical Computation. 



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CHAPTER XI 
POLES, POLARS, AND DIAMETERS 

168. Harmonic ratio. — If two points A and B divide a 
line segment M N externally and internally in ratios that 
have the same numerical values, then A and B are said to 
divide MN harmonically. A and B are called harmonic 
conjugates with respect to the line segment MN. 

Theorem. — If the points A and B, divide the line segment 
MN harmonically , then the points M and N divide the line seg- 
ment AB harmonically. 



-e- 



M A N B 

Fig. 146. 

Proof— By hypothesis, j^-= - ^ 

MA AN 

Taking this proportion by alternation, ^=r= = ~"da?' 

Multiplying both sides of this equation by — 1, and replacing 
~MA by AM and —BN by NB gives the required proportion 
AM _ AN 
MB NB 

159. Poles and polars. — Definition. — If a line drawn through 
some point Pi is allowed to rotate about Pi while cutting a 
conic in the variable points M and N, then the locus of all 
points harmonically conjugate to Pi with respect to M and JVis 
called the polar of Pi with respect to the conic, and Pi is 
called the pole of the locus. 

206 



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§159] 



POLES, POLARS AND DIAMETERS 



207 



To find the equation of the polar of Pi with respect to an 
ellipse, suppose the ellipse in Fig. 147 is given in the standard 
form, 

a 2 + 6* 

If P 2 is the conjugate of Pi with respect to the ellipse and 
P1P2 is any line cutting the 
ellipse in the points M and N, 
then M and N are harmonic 
conjugates with respect to the 
line PiPj. 

If the coordinates of Pi are 
(xi, yi) and the coordinates 
of P 2 are (x 2 , y 2 ) ; then by [4] the 
coordinates of M are 





i y 














—z*f^ X 




\b 


pfs, 


**VT / 


■h V 


N<r~ 




O V 


J*k 





Fig. 147. 



i r%X\ + rix 2 j r 2 yi + ny* \ 
\ ri + r 2 ' . ri + r 2 / 



and of N are 



( r 2 3i — T&2 
r\ — r 2 



ny\ 



ri - r 2 / 



Since M and N are points on the ellipse, their coordinates 
must satisfy the equation of the ellipse, therefore 



/ r 2 si + r x xt \ 2 
V n + r 2 / 



+ 



/ r 2 yi + riy 2 \ 2 
\ ri + r 2 / 



and 



/ r 2 a?i - r x x% \ 2 / r 2 g/i - rig/ 2 \ 
\ ri — r 2 / , \ ri — r 2 / 



+ 



= 1, 



- 1. 



o» 6 2 

Clearing each equation of fractions and subtracting the 
second from the first gives the equation 

4b 2 riT2XiX2 + 4a 2 rir 2 |/ij/ 2 = 4rir 2 a 2 6 2 . 
Dividing both sides of the equation by 4rir 2 and dropping 



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208 ANALYTIC GEOMETRY (§159 

the subscripts for the coordinates of the point ,Pj gives the 
equation of the polar of Pi 

b l X\X + a*yiy = a 2 6 2 . 

' ' a* + 6 2 " x ' 

This shows that the polar of a point with respect to an 
ellipse is a straight line. If the point Pi is outside the ellipse 
the line drawn through Pi will not always intersect the ellipse. 
Algebraically the points of intersection of such a line and 
the ellipse have imaginary coordinates, but the coordinates 
of the point conjugate to Pi with respect to these points 
with imaginary coordinates are real. Hence that part of the 
locus obtained outside of the ellipse is also included as part 
of the locus. 

In like manner it can be shown that the polar of a point 
Pi with respect to the hyperbola 

a 2 b 2 l 
is 

XiX y x y _ t 
a 2 b 2 " l ' 
Also the polar of Pi with respect to the parabola 

y 2 = 2px 
is 

y\y = px + pxi. 
Likewise the polar of a point Pi with respect to the general 
conic Ax 2 + Bxy + Cy* + Dx + Ey + F = is 

[45] AxiX + |xiy+|xyi + Cyiy + 5 x + ^xi+?y + 

^yi + F-0. 

The similarity should be noticed between this equation 
and the general equation of the conic written 

n in T\ 7} Jp w 

Axx +2 xy +"2 xy + C yy+2 X + 2 X+ 2 y + 2 y+Fts °' 



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§160] POLES, POLARS, AND DIAMETERS 209 

These equations show that the polar of a point Pi with 
respect to any conic is a straight line. 

160. Properties of poles and polars. — Theorem 1. — If two 
points are so situated thai one lies on the polar of the second, 
the second lies on the polar of the first. 

x 2 t/ 2 
Suppose the conic is the ellipse -\ + r, = 1, and the point 

is Pi(xi, yj. 

Then the pokrofPiis^ + ^ - 1. 

If P2 lies on the polar of Pi its coordinates will satisfy that 
equation, hence 

a 2 - 1 " 6* l ' 
But this is precisely the condition that Pi shall satisfy the 
equation 

x& y*y _ , 

a 2 "*" 6 2 " l ' 

which is the equation of the polar of P 2 . 

This proof can easily be extended to the general equation 
of the second degree. 

Theorem 2. — If tangents can be drawn from a point to a 
conic, the polar of this point passes through the points of context 
of the tangents. 

In Fig. 147, the tangent Pi22 meets the conic in two co- 
incident points at R. Since the conjugate to Pi lies between 
these two coincident points, it must coincide with R. Likewise 
the polar must pass through S. 

Theorem 3. — Tangents to a conic at the points where a line 
cuts the conic pass through the pole of the line. 

This follows at once from theorem 2 in conjunction with 
the assumption that only one tangent line can be drawn to 
a conic at a given point. 

Example. — Find the pole of the line x + 2y = 1 with respect to the 
conic 3s* + 4y* — 6. 

14 



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210 ANALYTIC GEOMETRY [§161 

The polar of the point (x h y x ) with respect to the ellipse Zx* + 4y* = 6 
is Zx x x + 4y\y = 6. 

Since this equation and x -\- 2y — 1 are equations of the same line the 
coefficients of x, y, and the constant term must be proportional, then 

TT and TT 

Hence X\ *= 2, y\ = 3, and the required pole is the point (2, 3). 

EXERCISES 

Find the equations of the polars of the points in exercises 1-8 with 
respect to the conies following. 

1. (2, 3) 3s* + 4y* = 6. 

2. (-1,6) 2s* + y* = -3. 

3. (-1,2) 3s* -2y* = 1. 

4. (1, -3) 2s* - 4y* - -5. 
6. (1, 2) y* - 6s. 

6. (-3, -2) s* = 4y. 

7. (1, 2) s* - xy + 2/* - 6s - Zy + 2 - 0. 

8. (3, -4) xy + Zy* + 3s + 1y + 1 = 0. 

Find the coordinates of the poles of the lines in exercises 9-16 with 
respect to the conies following. 

9. 2s + 4y = 1, 6s* + 4y* - 3. 

10. 2s + 2y - 1 - 0, 2s* + by 1 - 5. 

11. 2s - Zy - 6 - 0, 4s* - Zy 2 = 12. 

12. s - 2y + 4 = 0, y* - 2s - 0. 

13. s + y + 1 = 0, s* + Qy = 0. 

14. 4s + 5y = 2, s* + sy + y* - 3. 

16. 3s + by + 2 - 0, s* + 2y* + s + y - 0. 

16. 2s + 1 = 0, s* + 2sy + 2y - 2 - 0. 

17. Find a point which with* (2, 4) divides the line joining (1, 1) to 
(4, 10) harmonically. 

18. Prove that in any conic, the polar of the focus is the directrix. 

161. Diameters of an ellipse. — Definition. — The locus of the 
middle points of a set of parallel chords of a conic is called a 
diameter of the conic. 

To find the diameter of an ellipse, let its equation be given 
in the form of [32] and suppose that the slope of the parallel 
chords is mi. Unless m\ is infinite, the equations of these 
chords have the form y = m\X + c, where mi is constant for 



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§161] POLES, POLARS, AND DIAMETERS 211 

any one system of parallel chords, but c will have different 
values for different chords of the system. 

Suppose MN, Fig. 148, is one of these chords, the coordi- 
nates of M and N can be found by f 
solving simultaneously the equations 
y = m\X + c 

x t y 2 ' t\\\>Vft\Yl ** 

and Si + gi- h 

Eliminating y between these equa- 
tions gives Fig. 148. 

(6 2 + a*mi 2 )x* + 2a 2 cm x x + a 2 c 2 - a 2 b 2 = 0. 
The two roots of this equation are the abscissas of the 
points M and N. Half their sum is the abscissa of Pi(xt,y*), 
the middle point of MN. 

By a well-known theorem, Art. 4, the sum of the roots of 
the quadratic equation 

Ax 2 + Bx + C = 
S 
is equal to — -r- 

Hence a 2 cmi 

6 2 + a 2 m! 2 ' 

To find t/2, substitute this value of x% in the equation 

y = mix + c. 

Then y. = 6 Tqrjfi^- 

The relation between x% and y% for any one, and therefore 
for every one, of these parallel chords must be independent 
of c. Hence eliminate c by dividing y% by x^. This gives 

Hi = ft 2 

Xi a 2 mi 

Dropping the subscripts for P 2 gives the following equation 

of the diameter which bisects all chords of slope mil 

b 2 
y = — 5 — x* 

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212 ANALYTIC GEOMETRY [§162 

This is the equation of a straight line passing through the 
center of the ellipse. 

If mi is infinite, the parallel chords are all parallel to the 
y-axis and the symmetry of the ellipse shows the x-axis to be 
the diameter. 

If mi = 0, the parallel chords are all parallel to the z-axis, 
and the symmetry of the ellipse shows the y-axis to be the 
diameter. 

Since mi can have any value, any line passing through 
the center of the ellipse is a diameter. 

The length of a diameter of an ellipse is the distance be- 
tween the points where the diameter cuts the ellipse. 

162. Conjugate diameters of an ellipse. — The slope of the 
diameter bisecting all chords parallel to the diameter 



is 



-6* 



\a 2 m\I 



and its equation is 

y = mix (2) 

But the diameter (2) is the diameter of the ellipse parallel 
to the set of parallel chords of article 161. Hence the diameter 
(1) bisects all chords parallel to diameter (2), and the diam- 
eter (2) bisects all chords parallel to diameter (1). 

Two diameters such that each bisects all chords parallel 
to the other are called conjugate diameters. Hence diam- 
eters (1) and (2) are conjugate diameters. 

If m 2 is the slope of (1) 

6 2 

or 

6 2 

mim2 = — --=• 

a' 



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§163] 



POLES, POLARS, AND DIAMETERS 



213 



163. Diameters and conjugate diameters of an hyper- 
bola. — Methods exactly similar to those in articles 161 
and 162 show that all diameters of an hyperbola pass through 

X 2 V 2 

its center. The diameter of — 9 — j- 9 = 1, which bisects all 

a 2 b 2 ' 

chords of slope mi, Fig. 149, is 



V = 



b 2 




+*x 



Fia. 149. 



The slopes of two conjugate 
diameters of an hyperbola are 
connected by the relations 

b 2 

ar 

The length of a diameter of an hyperbola when the 
diameter meets the hyperbola is the distance between the 
points where the diameter cuts the hyperbola. If the di- 
ameter does not cut the hyperbola, its length is defined as the 

distance between the points where it 
cuts the conjugate hyperbola. 

164. Diameters and conjugate 
diameters of a parabola. — Let the 
slope of the parallel chords be mi, 
Fig. 150, and let their equations be 
y = m\X + c, where c will have 
different values for different chords. 

The ordinates of the points of in- 
tersection of the parabola y 2 = 2px, 
and these parallel chords are given by the equation 

m l2 / 2 - 2py + 2pc = 0. 

If y% is the ordinate of any one of their middle points 

V 

V 2 = ™ 
mi 

Since this equation is independent of c, it is the condition 



Fia. 150. 



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214 ANALYTIC GEOMETRY [§165 

that all points on the diameter must satisfy. Hence dropping 
subscripts, the equation of the diameter of a parabola is 

V 

This shows that the diameter of a parabola is a straight 
line parallel to its axis. Since mi may have any value except 
0, any line parallel to the axis of a parabola is a diameter. 

As mi approaches 0, the system of parallel chords approaches 
parallelism to the axis of the parabola, and y increases without 
limit. Hence the diameter bisecting chords parallel to the 
axis of a parabola does not he in the finite part of the plane. 

165. Diameters and conjugate diameters of the general 
conic. — Since the slope of a line remains unchanged by trans- 
lation of axes, all the results obtained so far hold good after 
translation for conies whose axes are parallel to the coordinate 
axes, providing that in the ellipse and hyperbola the major 
and the transverse axis respectively, and in the parabola the 
axis of the parabola are parallel to the x-axis. 

Formulas obtained for conjugate diameters, and equations 
of diameters do not hold true for rotation of axes unless 
account is taken in mi and m 2 of the change made by the 
rotation. 

EXERCISES 

1. Find the equation of the diameter of the ellipse 3s* + 4y* = 6, 
which bisects chords of slope 3. Chords of slope — J. 

2. Find the equation of the diameter of the hyperbola 2x* — 4y* = l t 
which bisects chords of slope 3. Chords of slope — J. 

3. Find the equation of the diameter of the parabola y* — 4z, which 
bisects chords of slope 3. Chords of slope — J. 

4. Find the equation of the diameter which bisects chords of slope 3, 
for the ellipse 2x* + Zy* - 4x - 12y + 2 = 0. 

Suggestion. — Translate axes to center of conic, and then translate 
back to the original axes. 

6. Find the equation of the diameter which bisects chords of slope 3, 
for the hyperbola 2s* - 3y* - 4a; + 12y - 22 - 0. 



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§165] POLES, POLARS, AND DIAMETERS 215 

6. If (2, 1) is one extremity of a diameter of the ellipse 4a?* + 9y* = 25, 
find the coordinates of the extremities of the conjugate diameter. 

7. If the point (1, 2) is one extremity of a diameter of the hyper- 
bola 25s* — 4y* = 9, find the coordinates of the extremities of the con- 
jugate diameter. 

x s y* 

8. If (x\, yi) is an extremity of a diameter of the ellipse ~i + m = 1, 

what are the coordinates of the extremities of the conjugate diameter? 

9. Prove that the sum of the squares of any two semi-conjugate 
diameters of an ellipse is constant and equal to a 1 + b*. 

x 2 v* 

10. If (xi, yi) is an extremity of a diameter of the hyperbola -^ — jk = 1, 

what are the coordinates of the extremities of the conjugate diameter? 

11. Prove that the difference of the squares of any two semi-conjugate 
diameters of an hyperbola is constant and equal to a* — 6*. 

12. Find the equation of the chord of the hyperbola 2a?* — Zy* = 6, 
through the point (4, 1) which is bisected by the diameter y = 4x. 

13. Find the equation of the chord of the ellipse x* + 2y* = 4, through 
the point (6, 3) which is bisected by the diameter Zy + x - 0. 

14. Find the equation of the chord of the parabola y % = 4x through 
the point (1, 6) which is bisected by the diameter y — 3. 

15. Prove that the polar of any point Pi(a?i, y\) on a diameter of an 
ellipse is parallel to the conjugate diameter. 

16. Two lines connecting a point on an ellipse with the ends of a 
diameter are called supplemental chords. Prove that supplemental 
chords are always parallel to a pair of conjugate diameters. 

17. Prove that if a parallelogram is inscribed in an ellipse its sides are 
parallel to conjugate diameters. 

18. Find the locus of the middle points of chords which connect the 
ends of pairs of conjugate diameters of a fixed ellipse. 



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CHAPTER XII 
ELEMENTS OF CALCULUS 

166, Introductory remarks. — As has been stated, the 
discovery of the methods of analytic geometry during the 
first half of the seventeenth century gave the first great start 
in the development of modern mathematics. During the 
latter half of the same century Newton and Leibniz, building 
upon the writing and teaching of Isaac Barrow and others, 
discovered the method of the infinitesimal calculus. In this 
subject are studied very powerful methods of investigating 
functions and problems concerning variables. It is in the 
calculus that we find the greatest development of mathe- 
matical analysis and its applications in almost every field of 
science and engineering. Some of these methods and 
applications will now be considered. 

Here, as is always the case in the study of mathematics, it 
is necessary to understand clearly what is under consideration 
and how it is represented in mathematical symbols. 

167. Functions, variables, increments. — Example 1. — If a 
suspended coiled wire spring has a weight attached to its 
lower end, the spring will be stretched. The amount of 
stretching will depend upon the weight, the greater the weight 
the greater the elongation. The elongation is then a function 
of the weight. If the weight is not so great that the elastic, 
limit of the spring is exceeded, the elongation varies directly 
as the weight The law connecting the variables is then stated 
by the linear equation 

V = kz, 

where y is the elongation, x the weight, and k a constant. 

216 , 



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§167] 



ELEMENTS OF CALCULUS 



217 



That is, y is a function of x, and a change in the variable x 
produces a corresponding change in y. 

A change in the weight is called an increment of the weight, 
or an increment of x, and is represented by the symbol Ax 
(read "increment of x yy or "delta x"). A corresponding 
change in the elongation is called an increment of the 
elongation, or an increment of y, and is represented by Ay. 

Here x represents the independent variable and y the de- 
pendent variable. 

It is evident that for every Ax there is a Ay. Their relation 
may be shown as follows: 



(1) 
(2) 




Ax N 



Fig. 151. 



For any particular value of x as x x , y x = kx x . 

If x = Xi + Ax, y x + Ay = k(x x + Ax). 

Subtracting (1) from (2), Ay = kAx. 

That is, Ay varies directly as Ax, 
and is independent of the value of x. 

This is shown graphically in Fig. 
151. The locus of y = kx is a 
straight line with slope k. Pi is a 
point on the line with coordinates 
(xi, yi). MN = PiQ = Ax, and 
QR - Ay. 

No matter what the magnitude of Ax, 

Ay = Ax tan QPiR = kAx. 

Example 2. — The distance s that a heavy body near the 
earth's surface falls from rfest in time t is given by the formula 

* 8 = hgt 2 . 
If t - t l9 Sl = \gt]. , (1) 

If t = h + A*, 8i + As = Mh+ A*) 2 . (2) 

Subtracting (1) from (2), As = ^(2*^+ At 2 ). 
That is, the value of As depends upon both t and At. 
This is shown graphically in Fig. 152. The locus of s = igt 2 
is a parabola. The point Pi has coordinates (h, si), and 



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218 ANALYTIC GEOMETRY [§167 

At and As are as shown in the figure. It is evident from the 
figure that As depends upon both t and At. 

Definitions and notation. — If y is a function of x it may be 
written y = f(z), which is to be read "y equals a function 
'y equals / of a." For convenience other symbols 
may be used for functions, as F(x), 
<p(x), /'(*)> etc. 

In the equation y = f(x), that is, 
when the equation expresses y explicitly 
in terms of x, y is an explicit function of x. 
If two variables are involved in an 
j+t equation in such a manner that it is 
necessary to solve the equation in order 
to express either explicitly in terms of the 
other, then either variable is said to be an implicit function 
of the other. 

Thus, in x 2 + y* = r*, y is an implicit function of x and x is an implicit 
function of y. 

If this is solved for y, y = ±\/r 2 — x 2 , in which y is an explicit func- 
tion of x. 

If solved for x, x «■ ±\Zr* — y 1 , in which x is an explicit function 
of y. 

Implicit functions of x and y may be written /(x, y), F{x, y), 
ip(x y y), etc. 

In the same discussion or problem the same functional symbol 
is used to represent the same function. 

Thus, if f(x) - 2x* + Zx + 1, 
then /(a) = 2a* + 3a + 1, 

and /(3) = 23* + 33 + 1= 28. 

If /(*> y) = 3s* + 4xy - y, 
then /(2, 3) - 3-2* + 4-2-3 - 3 = 33, 
and f(y, x) = Zy* + \xy — x. 

EXERCISES 

1. If y = 10g and x x is any particular value of x, find Ay when x 
takes the increment As. Find Ay when Xi = 4 and Ax = 2. Find Ay 
for any other value of as and Az = 2. Plot so as to show these graphically, 



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§167] ELEMENTS OF CALCULUS 219 

2. If y — 2x* + 1 and x x «■ 2, find Ay when As = 0.5. Find Ay when 
Ax - 0.01. Plot. 

3. Express the area A of a square as a function of its side x. Find 
AA for x = 6 and Ax = 1. Illustrate by means of a square. 

4. Express the area A of a circle as a function of its radius x. Find 
AA f or x — 10 and Ax =0.5. Illustrate by means of a circle. 

6. Express the area of a square as a function of its diagonal. Express 
its diagonal as a function of its area. 

6. Express the circumference of a circle as a function of its area. 
Express the surface of a sphere as a function of its volume. 

7. Express the volume of a right circular cylinder as a function of 
its radius and altitude. Express the altitude as a function of its volume 
and radius. Express its lateral area as a function of its volume and 
diameter. 

8. If /(*) - x* + 3s»- 2x - 4, find /(0), /»),/( -4). 

9. If F{x) - y/x^+T, find F(0), F(-3), F(W$). 

10. If <p{x) - logic x, find *>(100), ?(47.62) f ^(0.012). 

11. If /(*) - cos *, find /(30°), /(**•), /(240°). 

12. If/(s,y) - Zx*y + 4sy* - 2y\ find/(-s,y),/(ar, -y)./(-s, -y). 

13. If /(y) - 3*, prove /(s)-/(y) -/(* + y). 

14. If/(x) = sinsandFfc) = cos a;, prove that 

/(*+y) =/(^(y)+F(x)/(y). 

15. If y = sin a;, express x explicitly in terms of y. If y = 2*, express 
x explicitly in terms of y. 

In each of the following equations express each variable explicitly in 
terms of the other, if it can be done by methods previously studied. 

16. ~ + £= 1. 23. . ToTXTl - e ~*'- 
a* b 2 sin (2t + i*-) 

17. a* + y* - a*. 24. *>* cos 20 - a*. 

18. as* + y* = a*. 25. ^ sin tan $ = 4a. 

19. 3(3 — 2a)* — ay* = 0. 26. *>* cos $ = a 2 sin 30. 

20. «y + 4s 4 = 16. 27. logios— logi y+ 3 log™ a =0. 

21. 4x* + y* - 8x - 2y + 1 = 0. 28. sin" 1 x - sin" 1 y = 45°. 

22 *V _ /„ + 2 )t 29 * V - *~ 1 

22 * 16 - y* - {y + 2) * 39 ' *« - 1 - * + y 

30. If 8 - 16**and*i - 2, find As and -^ when A* - 1; when A* - 0.1; 

As 
when At = 0.01: when At - 0.001. What value does -tt seem to be 

' At 

approaching as At becomes smaller? 

At/ 
3L If y = x % and X\ = 1, find Ay and -^ when Ax = 10; when As = 1; 



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220 ANALYTIC GEOMETRY [J168 

when As — 0.1; when Ax — 0.01; when Ax — 0.001. What value does 
-~ approach? What then is the slope of the tangent at the point where 
x — 1 of the curve of y = x»? Plot. 

LIMITS 

168. Illustrations and definitions. — Considerable use has 
been made of limits in elementary geometry, trigonometry, 
and algebra, but much greater use is necessary in the study 
of calculus. The following are simple examples of limits: 

(1) The variable which takes the successive values 1.3, 1.33, 
1.333, • • • has as a limit \\. That is, the more figures 
there are taken, the more nearly the number approaches 1$. 

(2) The number \/2 is the limit of the successive values 
1.4, 1.41, 1.414, 1.4142, • • . The diagonal of a unit 
square is the limit of the line lengths represented by this 
series of numbers. 

(3) If a point starts at the end A of the line AB, Fig. 153, 
and during the first second moves half the length of the line 

to C; during the next second, half 

— | — |»m of the remaining distance to D; 

continuing in this way to move 

Fia. 153. i ii. ,i . . ,. , i 

half the remaining distance dur- 
ing each successive second, then the distance that the point 
is from A is a variable of which AB is the limit. 

12 

(4) If y = — -t-^ and z is a variable approaching 2 as a limit, 

then evidently y is a variable approaching 3 as a limit. 

Definitions. — When a variable changes in such a manner 
that its successive values approach a constant so nearly that 
the difference between the constant and the variable becomes 
and remains less, in absolute value, than any assigned posi- 
tive number, however small, the constant is the limit of the 
variable. 

The variable is also said to approach the constant as a limit. 
If the variable is represented by x and the constant by a, then 



4- 



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§169] ELEMENTS OF CALCULUS 221 

the statement "x approaches a as a limit " is written thus, 
x = a. 

lim 

The form ^^^ [f(x)] = A is read "the limit of f(x) as x 

X "~ a 

approaches a as a limit is A." 

When a variable changes in such a manner that it becomes 
and remains greater than any assigned positive number, 
however great, it is said to increase without limit or to become 
infinite. 

The notation to represent this is x = », which is read 
"x increases without limit" or "x becomes infinite." 

lim 

The form „_ m [/(*)] = A is read "the limit of fix) as 

x becomes infinite is A" 

169. Elementary theorems of limits. — The following theo- 
rems will be found useful in dealing with limits. They are 
given here without proof. 

• (1) If two variables thai approach limits are equal for all their 
successive values, their limits are equal. 

(2) The limit of the sum of a constant and a variable that 
approaches a limit is the sum of the constant and the limit of 
the variable. 

(3) The limit of the produdt of a constant and a variable that 
approaches a limit is the product of the constant and the limit of 
the variable. 

(4) If each of a finite number of variables approaches a limit, 
the limit of their sum is the sum of their respective limits. 

(5) If each of a finite number of variables approaches a limit, 
the limit of their product is the product of their respective limits. 

(6) // each of two variables approaches a limit, the limit of 
their quotient is the quotient of their limits, except when the 
limit of the divisor is zero. 

If the limit of the divisor is zero the limit of the quotient 
may have a definite finite value or the quotient may become 
infinite, but it is not determined by finding the quotient of 



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222 ANALYTIC GEOMETRY [§170 

the limits of the two variables. The calculus determines such 
limits as these exceptional cases. 

170. Derivatives. — The fundamental conception of differ- 
ential calculus, and one that is of the greatest importance in 
mathematics, is the derivative of a function. Using the 
notation of this chapter the derivative is defined to be the 

At/ 
limit approached by the quotient ^ as Ax approaches zero. 

If the curve, Fig. 154, represents the function y = /0r), 

Ay 
the quotient ^- is the slope of the secant line PiP„ 

If Pi remains fixed and Ax approaches zero as a limit, 

the point P moves along the curve 
and approaches Pi as a limit, and 
the secant PiP turns about Pi to 
the limiting position QR, which is 
defined to be the tangent to the 
curve at the point Pi. 

Hence, the slope of (he tangent is 
precisely the quantity called the 
Fig. 154. derivative. 

It is evident that the value of the derivative depends upon 
the position of Pi on the curve. 

Definition. — The slope of a curve at any point is the slope of 
the tangent to the curve at that point. 

du 
The notation for the derivative is j-> read "the derivative 

of y with respect to x." Then by definition 
dy _ lim rAy~| 
dx~ Ax=0 LAxJ 
Of course, the independent variable and the function may 

du iim rAtn 

be represented by other letters. Thus, -37 = a*^o I aFJ 




_. . dy 

The notation -f- 

dx 



du 
is used to indicate the value of j- for 

x—xi ax 



the particular value X\ of x. 

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§171] 



ELEMENTS OF CALCULUS 



223 



Example. — Given y = x 2 , find 



dy 



and thus find the slope of the 



£-*+**• 



dx\x—2 1 

tangent to the parabola at the point (2, 4). Also find the equation of 
this tangent and plot. 

Solution. — (1) Given y = x 2 . 
. (2) When x - 2, y - 4. 

(3) If a; takes an increment As, y + Ay = (2 + Ax) 2 = 4 + 4Ax + A?. 

(4) Subtracting (2) from (3), Ay - 4Ax + Ax 2 . 

(5) Dividing by Ax, 

(6) Letting Ax = 0, 5i|x = 2 ~ 4 * 

Hence the slope of the tangent to the para- 
bola at the point (2, 4) is 4. 

The equation of this tangent by [15] is 

y — 4 = 4(x — 2), or 4x — y = 4. 

The plotting is shown in Fig. 155. 

171. Tangents and normals. — It fol- 
lows from the preceding article and [15] 
that the equation of the tangent to the 
curve y = f(x) at the point {x ly j/i) is 



[46] 



dy 
y - yi= di 



X-Xi 



(x - Xl ). 




+X 



Fig. 155. 



Definition. — The normal to a curve at any point is the line 
perpendicular to the tangent to the curve at that point. 

Then by [9] and [15] the equation of the normal to the curve 
V — f( x ) a * the point (x h j/i) is 

[47] . y-y 1 =-_l_(x-X 1 ). 



dy 
dxx=xi 

Example. — Find the equations of the tangent and normal to the ellipse 
4x 2 + 9y 2 = 36 at the point (xi, y{). Also find these equations when 
xi = 2. Plot. 

Solution.— (1) Given 4x 2 + 9y 2 = 36. 

(2) Let x = xi and y = y u 4xi 2 + 9yi 2 = 36. 

If x takes the increment Ax, y will have the increment Ay, and 
' (3) 4(xj + Ax) 2 + 9(2/i + Ay) 2 - 36, _ 

or 4xi* + 8xiAx + 4Ax* + W + lSyiAy + 9Ay* = 36. 



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224 



ANALYTIC GEOMETRY 



[§171 



(4) Subtracting (2) from (3), 8siAa: + 4A3* + ISyiAy + 9Ay* - 0. 



(5) Transposing and arranging, ~ = 



Sx t +4As 



18yi + 9Ay 

Passing to the limits and noticing that Ay * as Ax * 0, 
dyl __ 4a?i 

Substituting in [46], the equation of the tangent is 



y -Vi 



4«i / \ 



or 4x& + 9yiy - 4*1* + 9yi 2 . 

Since by (2) 4»i f + 9yi f — 36 the equation of the tangent is 

4*1* + 9y# - 36. 
Similarly the equation of the normal is 

y ~ Vl " te (X " Xl) ' 
When a; - 2, y - ±f V5- 



J\W4V») 




^Jf 



(2.-^^6) 



Fig. 156. 



Substituting these values for xi and y\ in the equation of the tangent, 
the equation of the tangent at (2, f y/b) is 

4-2x + 9-fVoV - 36, or 4s + 3\/% - 18. 

And the equation of the tangent at the point (2, — fv/5) is 
4-2* +9(- W&)y =36, 
or 4s - 3\/% - 18. 

Likewise the equations of the normals are, by [47]: 
at the point (2, K/5), 9Vpx - 12y - 10\/5; 
and at the point (2, ~f a/5), 9\/6s + 12y = 10V5. 

The plotting is shown in Fig. 156. 



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§172) ELEMENTS OF CALCULUS 225 

EXERCISES 

1. Given y — x* t compute the values of Ay and ~ when x « 0.5 and 

Ax « 1, 0.1, 0.01, and 0.001 respectively. 

2. Find the slope of the tangent and normal to y — 4x* at the point 
where z * 0.5. 

dy\ 

3. Given y — x* + 2, find -p — « and write the equation of the tan- 
gent and normal at this point. Check the result by plotting. 

Find -£■ at the point G&i, yi) for each of the following: 

4. y - 3s* - 1. 10. y - x* + 2s J + 5. 

5. y ■ - x* + 4. 11. y - 3x* - 4x* + Ox. 

6. y - 2x + 5. 12. x* + 2y» - 16. 

7. xy - 4. 13. 4x* - 9y» - 36. 

8. y* - 2px. 14. y* - 4x + 8. 
* 1 «. s + 1 

16. Given the parabola y* — 2px, find the equations of the lines tangent 
to the parabola at the extremities of the latus rectum, and show that 
they meet on the directrix. 

17* Find the slope of the circle x % + y* — 25 where x — 2, (a) when 
the point is in the first quadrant, and (6) when the point is in the fourth 
quadrant. 

18. Find the angle that the line 3x — 4y + 7 » makes with the 
circle x* + y* — 25 at their point of intersection in the first quadrant. 

19. At what angle does the circle x* + y* = 16 intersect the circle 
x f + y f = 8x at their point of intersection in the first quadrant? 

ALGEBRAIC FUNCTIONS 

172. Differentiation by rules. — The process of finding the 
derivative of a function is called differentiation. 

The method used in the preceding articles in finding the 
derivative is? called the fundamental method since it is based 
directly upon the definition of a derivative. The derivative 
of any function can be found by this method, but the work 
can be greatly shortened by using rules or formulas which 
can be established by' fundamental methods or otherwise. 
The rules needed in differentiating algebraic functions will be 



15 



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226 ANALYTIC GEOMETRY [§173 

considered first, and later some of those necessary to 
differentiate trigonometric, exponential, and logarithmic 
functions. 

In the formulas, x, y, u, and v denote variables, which, of 
course, may be functions of variables, and a, c, and n denote 
constants. 

173. The derivative when f (x) is x. — Since the equation 
y = x represents a straight line with slope equal to 1, and by 

dy 
article 170, -£ is the slope of the curve at any point, it follows 

that 

I <* = ! 

dx 

In general, the derivative of a variable with respect to itself is 
unity. 

174. The derivative when f (x) is c. — Since y = c is the equa- 
tion of a straight line with slope equal to 0, it follows that 

n. £ = o. 

dx 

In general, the derivative of a constant is zero. 

175. The derivative of the sum of functions. — Given 
y = u + v, where u and v are functions of x, and let Ay, Ati, 
and At; be the increments of y, u, and v } respectively, corre- 
sponding to the increment Ax. 

Let x = xi, then y\ = U\ + v\. 

Let x = Xi + Ax, then yi + Ay = u x + Aw + V\ + Av. 

Subtracting, Ay = Au + At;. 

Ay Au Av 
Dividing by As, aS = aS + A^' 

__ du i^HL 

X — Xi dX X = Xi dX X**X\ 

It is evident that any number of functions can be treated 
in a similar manner, then 



Let Ax = 0, then -/ 

' dx 



ttt d(u + v + w+ • • -) _ du dv dw 
m * S " dx + dx "*" dx + 



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§176] 



ELEMENTS OF CALCULUS 



227 



Or, the derivative of the sum of any number of functions is 
equal to the sum of their derivatives. 

Example. — If y = x* + 3a;* — 4z + 3, 

dy = dM _ d(3s») _ d(4x) _ rf(3) 
dx dx dx dx dx 

176. The Derivative of the product of two functions. — 

With the notation as in the previous article, given y = uv. 
Let x = Xi, then j/i = urn. 

Let x = Xi + Ax, then yi + Ay = (t*i + Aw)(vi + At;). 
Subtracting, Ay = wiAt; + v\bu + Aw At;. 



Dividing by Ax, 



Ay At; , Am , A At; 

a — ^i 7 V *>i 7 — hAw— • 

Ax Ax Ax Ax 

At; 
Let Ax = and notice that Au-^- also approaches zero as 



a limit, 
then 



IV. 



dy 
dx 



— dt; 
x=xi dx 



, du 
x=a;i dx 



X=*X\ 



d(uv) dv , du 



Or, #ie derivative of the product of two functions is equal 
to the first times the derivative of the second plus the second 
times the derivative of the first 



Example. — If y 
dy 
dx 



(x - 2)(s* + 1), 

S-b-t ^ + n + v + i)*^*!. 



dx 



dx 



177. The derivative of the product of a constant and a 
function. — Given y = cu, where c is a constant. By the 
previous article 



But 
V. 



dy __ du . dc 
dx dx dx 

dc 

— - = 0. 
dx 

d(cu) du 



By II. 



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228 ANALYTIC GEOMETRY [§178 

Or, the derivative of the product of a constant and a function 
is equal to the constant times the derivative of the function. 

Examplea.^-Ify - 4(s - 2), ^ - 4 d(x ~ 2) - 

T - • u dy ldu 

"a dx adx 

178. The derivative of the quotient of two functions.— Given 

u 

„-- 

Let x = xi, then yi = — • 

V\ 

Let x = xi + Ax, then yi + A# = l 



Subtracting, Ay = 



Vi + Av 
U\ + Am _ Wi _ yiAu — U\Av 
Vi + Av Vi "~ Vi(vi + Av) 



Aw At; 

Aj/ Ax Ax 



Dividing by Ax, ^ = ^^ ^ } 

du 



dy I r <ix 



dtl 



x*=xi ax \x**x\ 





«i» 


du 


dv 


V di 


dz 



Let Ax = 0, then , - , — 

ax \x=xi 

vi. -'•-¥-- 

dx v 2 

Or, the derivative of the quotient of two functions is equal to 
the denominator times the derivative of the numerator minus 
the numerator times the derivative of the denominator, all divided 
by the square of the denominalor. 

r, i re x — 1 dy dx dx 

179. The derivative of the power of a function. — Given 

y = u n . 

(a) When nis a positive integer. 

Writing as a product y = uu*~ l . 

r*. dy m .du , d(u H -*) ^ T - r - 



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§179] ELEMENTS OF CALCULUS 229 

Writing w*" 1 as the product uu*-*, 

ax ax 

When this process is performed n times the last term will 

d(u n ~ n ) 
contain — -3 — -> which is zero by II. 

vn. .-. ^) = » u - l J 1 . 

dx dz 

(6) When n is a fraction, — , toAere p and q are positive integers. 

p 
Given y = u« 

Raising both sides of the equation to the qth power, 

y q = u*. 

Then M 9 " 1 ^ = vuV ~ l ~T' By ( a ) of this article - 

SI" f ^ ^ — P***"" 1 ^ 
g da;' dx " gj/*" 1 dx 

©u^ 1 du p --1 du 

5.(8-1) dx g dx 

qu« 

dx g dx 

(c) TTfcen n ts negative, either integral or fractional. 
Let n = — m. 

Then y = w- m = — 

Clearing of fractions, yu m = 1. 

Then myu^ 1 ^ + w- ^ = 0. By IV, VII, and II. 



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230 ANALYTIC GEOMETRY [§180 

Solving for & £ = - *2^ ** 

ax ax u m ax 

dx dx 

k Therefore formula VII is established when the exponent is 
a positive or negative integer or fraction. It is expressed in 
the following rule: 

The derivative of a function affected by an exponent n is equal 
to n times the function affected by the exponent n — 1, times the 
derivative of the function. 

Examples.- -If y - (** + x + 1)*, ^ = 4(s« + x + l)' ^*' + * + 1} « 

180. Summary of formulas for algebraic functions. — 

The formulas here summarized enable one to differentiate 
algebraic functions. 

I - = 1 
x * dx l * 

n. £ = o. 

dx 

m d(u + v + w H ) __ du dv dw 

U1# dx ~~ dx "*" dx "*" dx + ' ' % 

TTy d(uv) dv . du 
IV. \ ' = u r + v-j — 
dx dx dx 

V - dx " c dx 



VI. 



o 



du dv 

dx dx 



dx v 2 

._ d(u w ) n .du 

VII. -~- = nu*- 1 ^ — 
dx dx 



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§181] ELEMENTS OF CALCULUS 231 

181. Examples of Differentiation. — If the formulas and 
rules of differentiation are well learned, their application is 
one of the easiest processes in mathematics. 

Example 1. — Given y = 7s 8 , find -p . 

| = «2g> = 7 ^g = 73x» | = 21*'. By V, VII, and I. 

dv 
Example 2. — Given y = x* + 2x % — 5s + 6, find -p. 

ds dx dx dx dx 

- 3s* + 4s - 5. By VII, V, II, and I. 

Example 3.— Given y - (s 2 + 2s) (3s - 2), find-^. 

g = (x , + !te) ^l2) +( 3 ;c _ 2) ^!^) By IV. 

- (s 2 + 2s)3 + (3s - 2)(2s + 2) - 9s* + 8s - 4. 

By I, II, III, V, VII. 

Example 4. — Given y ■» = — — r> find -X 

• (3,-1)^^ -(,.+2)^^ 

<*y_ ' *c dx B VI 

ds " (3s - l)i * y V1 - 

_ (3s - l)(2s) - (s 2 + 2)3 _ 3s 2 - 2s - 6 



(3s - 1)* ~ (3s - 1)* 

Example 5. — Given y = ^x 2 + 3s, find ~. 

dy _ d^x* + 3s _ d(s 2 + 3s)* _ ... , d(s 2 + 3s) 

ds S di i{X +6X) di 

- *(*» + 3s)~*(2s + 3) .- -j^= • 
3V (s* + 3s) 2 

EXERCISES 

In the following find the derivative of the function with respect to 
the independent variable. 

1. y - 3s 2 . 6. y = 4 Vs. 11. y = -17yV8. 

2. y - 5s*. 7. y - 3-^s. 12. y = -2^/s 4 . 

3. y = 7s*. 8. y - -^s"». 13. « - igt*. 

4. y = as*. 9. y - 3s"*. 14. « = 4*.' 

5. y = fs*. 10. y - -4s*. 16. 8 = JA 
16. y =» s 4 + 3s 2 -f 2. 17. y - 3s 2 - 2s + 6. 



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232 ANALYTIC GEOMETRY [§181 

18. y - x« - x» + 3*. 22. y - (3s* + 2)« - 2s. 

19. y - a* - ar* + 4. 28. y - (2x + 3)* - 3s. 

20. y - x* - 3ar« + 2. 24. y - \/2x» - 7x. 

21. y - (2x + 1)' - 3. 25. y - -^x« + 7x - 2. 

26. „--,. 29. f-jqpj- 82. f-^jrj- 



2x« Q * # * (x« - 1)* ™_*_ a; 8 + 1 

85. y - 3a* - 4x • + 3a; 4 - 3. 48. « - VT+T + <^2* - 3. 

86. y - Vx~+~1 - Va^T- 44. 8 - t* + 2T* + 3*«. 

37. y - V3a; 8 + 7x* - 3a; + 2. 45. y - (a; 8 + l)(x* - 2a; + 1). 

88. y - Vax* + bx +_c - Vx~+"S- 46. y « (x + a)»(x - 6)» 

39. y - x»(x* + 5)*. 47. y - (a; + l)«(2x - 1)«. 

2a; - 1 AO 2x* - 1 

51. Find the slope of the tangent line to the curve y — a;* at the point 
where x = 0. At the point where x = 1. Where x ■» 2. 

52. In exercise 51, what is the slope of the curve at each of the points? 
How many times faster is y increasing than x at each of the points? 

53. If a point is moving from the origin along the curve y — 2x* 
in the first quadrant, what is the relative rate of increase of x and y 
when x = 1, 2, and 4, respectively? 

54. Find the equations of the tangent and the normal to the curve 
y = x 9 + 4a; 1 + x - 6 at the point (0, - 6). At the point (2, 20). 

55. In the curve of exercise 54, where is the tangent line parallel 
to the x-axis? 

56. Find the equations of the tangent and the normal to the curve 

y = x + -j at the point (xi, yj. 

57. Find the point on the curve y — x* + 3x* — 4x — 12 at which the 
tangent has a slope of — 7 V What is the equation of the tangent at this 
point? Plot the curve. 

58. At what angle does the line y = x — 1 intersect the parabola 
y* + 4x - 4? 



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§182] ELEMENTS OF CALCULUS 233 

x 9 

59. Show that the parabola y* — Aax and the cissoid y — s — ^T~ 

intersect at right angles at the origin. 

60. The heat H, required to raise a unit weight of water from Q°C. 
to a temperature f, is given by the formula 

H - t + 0.00002** + 0.0000003* 8 . 

JIT JTT 

Find -rr and compute the value of -=r- where t° — 35°C. 

182. Differentiation of implicit functions. — In the previous 
exercises, the dependent variable in each was expressed as an 
explicit function of the independent variable. Often it is 
either not convenient or not possible to express one variable 
as an explicit function of the other. In such a case the usual 1 
rules for finding the derivative can be applied and the desired 
derivative found as an implicit function of the variables 
involved. The method can be best illustrated by examples. 

Example 1. — Given x* + y % =» 25, find -p as an implicit function of 

x and y. 

Since y is a function of x, the left hand member is the sum of two 
functions of x. 

Differentiating, 2x + 2yj- =0. 

dy _ x 
dx y 

Example 2. — Find the equation of the tangent line to the curve 
x % — y h + x % — y « at the point (1, 1). 

Solution.— Differentiating, 5x 4 - 5y A -p + 3s 2 - ^ «0. 

Solving for -^t 

When x = 1 and y = 1, 

Then the slope of the tangent at (1, 1) is f . 

Hence the equation of the tangent is y — 1 — f (x — 1), 

oi 4x - 3y - 1 - 0. 



dy 


5x* + 3z J 


dx 
dy 

dx s 


5y* + 1 
4 
= 3. 



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234 ANALYTIC GEOMETRY [§182 

EXERCISES 

In the following find the derivatives as implicit functions. 
1. s' + 2,« = ^findg. 
• 2. y* + y = s* +.*, find ^. 

S.*+g-l f fad*. 

a* 6 2 ' dx 

4. pv = c, find -p and -r-« 

5. x* - 4*V + y» = 0, find g| and p. 
6.**+y* = o^, find g and g. 

7. x 1 + y 1 - a 1 , find ^[. 

8. (s + y)» + Or - y)» - a, findg. 

• t.(p+5)(.-»-MI-dJ M id* 

10. Find the equations of the tangent and the normal to the circle 
x* + y % « 25 at the point (3, 4). 

11. Find the equations of the tangent and the normal to the circle 
x* + y* - 4x + 6y - 24 = at the point (1, 3). 

12. Find the equations of the tangent and normal to the ellipse 
162 s + 25y* = 144 at the point in the first quadrant where x — 2. 

Show that the tangents to the following curves at the point (x h y t ) 
are as given. 

Equation of curve Equation o f tangent 

13. z* + y* = r*. xis + y# = r*. 
H. y 2 = 2px. yiy = p(s + Xi). 
15. a; 1 - 2py. XiX - p(y + yi). 

a 2 6 2 o f o f . 

a 2 6 2 = * 

18. rry = c. x x y + yix - 2c. 

19. Find the equations of the tangent and normal to the parabola 
3* + y* = a* at the point (*i, f/i). 



a 2 6 2 



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§183] 



ELEMENTS OF CALCULUS 



235 



FURTHER USES OF THE DERIVATIVE 

183. Discussion. — By methods of analytic geometry the 
properties of the locus that are most conveniently discussed 
are the intercepts, symmetry, and extent (See Art 43). By 
means of the derivative other properties may be discussed. 
Some of these will be considered in the following articles. 

The discussion will be confined to equations (1) whose 
curves have no break, at least in the part of the curve con- 
sidered; and (2) where for each value of the independent 
variable there is but one point on the curve. Such curves, 
as well as the functions giving rise to them, are said to be 
continuous and single-valued. 

184. Properties of a curve and its function. — If the curve, 
Fig. 157, is thought of as traced by a moving point passing 
from left to right, the following properties may be noted: 




(1) The curve is falling from A to B, from D to F, and from 
H to J; and the corresponding function is decreasing. 

(2) The curve is rising from B to D, from F to H, and from 
J to K; and the corresponding function is increasing. 

(3) If the curve rises to a certain position and then falls, 
such a position is called a maximum point of the curve. D 
and H are such points. The ordinate, that is, the value of 
the function, at such a point is called a maximum ordinate 
or maximum value of the function. 

(4) If the curve falls to a certain position and then rises, such 
a position is called a minimum point of the curve. B, F, find 



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236 ANALYTIC GEOMETRY [§186 

J are such points. The ordinate, that is, the value of the 
function, at such a point is called a minimum ordinate or a 
minimum value of the function. 

(5) The curve is concave upward between A and C, E and <?, 
and I and K. It is concave downward between C and E, and 
G and I. 

(6) Points C, E, G, and J where the concavity changes, are 
called points of inflection. 

Curves may have other peculiarities, but these will not be 
considered here. 

186. Curves rising or falling, functions increasing or 
decreasing. — Since by definition, Art. 170, the slope of a curve 
at any point is the pame as the slope of the tangent at that 
point, it follows that when the slope is positive the curve is 
rising, and when the slope is negative the curve is falling. 
This is, of course, when passing from left to right. 

Stated with reference to the function this becomes the 
following very useful principle: 

When the derivative of a function is positive, (he function 
increases as the independent variable increases; when (he derivar- 
tive is negative, the function decreases as the independent variable 
increases. 

It also follows that the ratio of the change of the function 
at any point to that of the variable is equal to the value of 
the derivative of the function with respect to the variable, for 
that point. 

Example 1. — For what values of a? is the curve y «■ z* rising and 
for what values falling? 
Solution. — Given y — x*. 

Then %. - 2*. 

dx 

Now 2x is positive when x is positive, and negative when x is negative. 
Hence the curve is rising when x>0, and falling when x<0. 

Example 2. — For what values of 3 is the function y = x* increasing 
and for what values decreasing? 

Here -¥ = 3a; 2 , which is not negative for any value of x. 



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§186] ELEMENTS OF CALCULUS 237 

Hence the function is never decreasing. 

Is it always increasing? 

Example 3. — For what values of 3 is the curve y — }*»— \x* — 6* 
rising and for what values falling? For what values of x is y increasing 
6 times as fast as x? 

Solution, — Given y — }»»— \x* — Ox. 

Then -^ - x % - x - 6. 

Factoring, ^ - (x + 2)(* - 3). 

Then -p is positive when x< — 2 and when »>3, and negative when 

-2<s<3. 

Hence the curve is rising when x<—2 and when x>3, and falling when 
-2<a;<3. 

The values of x for which y is increasing 6 times as fast as x can be 
found by putting x* — x — 6 — 6, and solving for x. 

This gives x — 4 or —3. 

EXERCISES 

Passing from left to right, for what values of x are the loci of the 
following equations rising and for what values falling? 

1. y = 3x — 6. 8. y - x* — x* — 2x. 

2. y - 4a; 1 + l&r - 7. 9. y - s» - 2a* + 3 - 3. 

3. y - VS. 10. y(l + a; 2 ) - a. 

4. y« - 8s*. 11. y(a* - 1)* - s». 

5. y - a* + 3. 12. 6y - 2s 8 - 3a* - 12s - 6. 

6. xy - 15. 13. y « x 4 — 6s 2 + &c + 6. 

7. y - *» - 9x. 14. y - (*« - 1)*. 

15. In exercise 8, how many times as rapidly as x is y increasing 
when x - 10? When x - 3? When 3 - -1? When x - 0? 

16. In exercise 9, for what values of a; is y increasing 7 times as rapidly 
as a;? For what values of x is y decreasing 4 times as rapidly as a; is 
increasing? 

186. Maximum and minimum. — From the definitions of 
article 184, it is clear that if a curve is plotted in rectangular 
codrdinates, the curve is rising at nearby points on the left of 
a maximum point, and falling at nearby points on the right. 
For a minimum point the curve is falling for nearby, points on 



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238 ANALYTIC GEOMETRY [§186 

the left and rising on the right. The student can readily 
state this with reference to the function. 

It is evident that at a maximum point or a minimum point 
like those shown in Fig. 157, the tangent line is parallel to 
the x-axis, that is, its slope is zero. 

It follows that these points can be determined from the 
f miction as follows: 

(1) Equate -j- to zero and solve for x. 

(2) Determine whether j- is positive or negative for nearby 

points on the left and right. 

dv dv 

A point where -r- = is a maximum point if -r >0 for 

dy 

nearby points on the left and -r- < f or nearby points on the 
right. 
A point where -r- = is a minimum point if -t- < for 

dii 
nearby points on the left and -r- > f or nearby points on the 

right. 

It is distinctly understood that these 
tests determine only such points as are 
illustrated in Fig. 157. For cusp maxi- 
mum and minimum points as shown in 
Fig. 158, the tangent is perpendicular 
Fig. 158. to the 3>a j^ 8 an( j hence -p =" « . 

Example. — Determine the maximum and minimum points of the func- 
tion y ■» x s — 3s f + 4 and plot the curve. 
Solution. — Given y - x* — 3x f -f 4. 

^ - 3s* - Ox - Zx{x - 2). 
ax 

. ' . ■¥ = f or x = 0, and x = 2. 
dx ' 

dv 
When x < but near 0> ~ > .0 and the curve is rising. 




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§187] 



ELEMENTS OF CALCULUS 



239 



When x > but near 0, -?■ < and the curve is falling. 

ax 

. * . the curve has a maximum point when x = 0. 

When x < 2 but near 2, -]*■'< and the curve is falling. 
ax 

When x > 2 but near 2, -p > and the curve is rising. 
. * . the curve has a minimum point when x — 2. 

Plotting.— When 3 - 0, y = 4. . * . (0, 4) is 
a maximum point. 

When x = 2, y = 0. . * . (2, 0) is a minimum 
point. 

Factoring, y = (a + 1)(* - 2)(s - 2). 

. * . the z-intercepts are —1, 2, and 2. 

,A few other points will make the plotting 
fairly accurate. See Fig. 159. 



X 


1 


3 


4 


- 2 


V 


2 


4 


20 


-16 




Fig. 159. 



EXERCISES 

Determine the maximum and minimum points of the following curves 
and plot. 

1. y - x*. 5. y - (x + 4)(s - 2)(* - 4). 

2. 2y - a 1 - 4* + 6. 6. y = s« - 7s* + 36. 

3. y » 6x - s« + 4. 7. 16y = a; 1 - 32x. 

4. 4s* + 9y* = 36. 8. y = s« - 4s«. 

9. By finding the maximum point of the curve, find the coordinates 
of the vertex of the parabola 2x* — l&r + 15y — 21 — 0. 

10. The equation of the path of a projectile is 



y = tan a-x — ; 



Art 94.) 



2v* cos* a ' 
Find the maximum height to which the projectile rises. 

187. Concavity and points of inflection. — It is evident 
from an inspection of a curve that is concave upward that the 
tangent line turns counter-clockwise in passing along a curve 
from left to right, that is, the slope of the tangent increases. 

Likewise, if the curve is concave downward, the tangent line 
turns clockwise, that is, the slope of the tangent is decreasing. 



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240 ANALYTIC GEOMETRY [§187 

Thus, in Fig. 160, the tangent line turns counter-clockwise in passing 
from A to Z>, and the slope increases from a negative value at A to a 
positive value at D. 

likewise, in Fig. 161, the tangent turns clockwise in passing from 
A to D, and the slope decreases from a positive, value at A to a negative 
value at D. 





Fig. 160. Fig. 161. 

It remains to determine how the concavity of a curve can be 
determined from its function. 

Since the derivative of a function of a: is itself a function 
of x, it is evident that the derivative of this first derivative 
may be found. It is called the second derivative of y with 
respect to x. 

If y = f(x) y the second derivative is jzviz) an d is 

d*y 
represented by the symbol -ry 

Thus, if y - x* — 6x* + \2x - 3. 

^ - 3s« - 12* + 12, 

and p{ - 6s - 12. 

dx* 

d*y 
From the foregoing, it is evident that when -r\ is positive, 

-i| is increasing; and when -r- t is negative, -j- is decreasing. 

Or, if y = f(x) is the equation of a curve, the slope of the 
tangent is increasing when passing from left to right and the 

dhi 
curve is concave upward for the values of x that make -t4 

positive. 



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§187] ELEMENTS OF CALCULUS 241 

Likewise, the curve is concave downward when -j-? is 

' ax 2 

negative. 

From (6) of article 184, it is evident that a point of inflection 
is a point on a curve at which the concavity changes from 
upward to downward or vice versa. A point of inflection can 

be determined by finding the values of x for which -r\ changes 

sign, providing the function is finite for that value of x. 

Example. — Investigate y ■» x* — 3x* + x + 2 for concavity and 
points of inflection. 

Solution. — Given y « x* — 3s* -f x + 2. 

P - &f - 6* + 1. 

g-te-6-6(*-l). 

Since when »<1, 6(a; — 1) is negative; and when «>1, 6(a; — 1) is 
positive, the curve is concave downward at the left of x = 1, and concave 
upward at the right of & = 1. Therefore, it has a point of inflection 
at the point (1, 1). 

EXERCISES 

In exercises 1-10 investigate for concavity and points of inflection. 

y - (a; + 2)(* - 2)(x - 3). 
y = 3s 4 - 4s 8 - 1. 
y = x* — 4a; 2 + 4x — 1. 
y = a; 4 - 2a; 2 + 40. 
y - 3a* - 16a; 8 - 6a; 8 + 48a; + 17. 

11. In the example, Art 187, find the slope of the tangent to the 
curve at the point of inflection, find the maximum and minimum points, 
and plot the curve. 

12. In the example referred to in exercise 11, if the curve is being 
traced by a point moving from left to right, for what values of x does 
y increase at the same rate as a;? How rapidly is the curve rising when 
x - 3 if x is increasing at the rate of 2 inches per second? 

13. Investigate the greatest possible number of points of inflection 
of the curves of 

(1) y = ax 1 + bx -f c, 

(2) y = ax* + bx % + ex + d. 

(3) y - ax 4 + bx* + ex 1 + dx + e. 

16 



1. y — a; 8 . 




6. 


2. y - x 4 . 




7. 


3. y « a* 




8. 


4. y = 3a; - 


-a; 8 . 


9. 


5. y « x A - 


-6a;». 


10. 



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242 



ANALYTIC GEOMETRY 



Hiss 



In exercises 14-19 plot the curves showing the values of y, -& and 

-z-^f using the same set of axes for the three curves of each. What 
facts can be read from these curves? 

14. y - 4s«. 17. y - (x + 2)(x - 2)(x - 3). 

15. y - 3x*. 18. y - x* - 12s + 7. 

16. y - 3x - x 9 . 19. y = *r 4 - 2x* - 8. 

DIFFERENTIALS 

188. Relations between increments. — When two variables 
are so related that the ratio of their corresponding increments 
is constant, either variable is said to change uniformly with 
respect to the other. 

When the variables are related by an equation of the first 

At/ 
degree, as y = mx + 6, where Ay = mAx, then ^- = m. 

That is, either variable changes uniformly with respect to the 
other. 




N 



M AXM- 



+-X 



AA 



C AXM 



Fig. 162. 



Fig. 163. 



This is also evident from Fig. 162, in which y : = mx + b is 
the equation of the line PP\ with slope m. P is any point op 

At/ 
this line and -r-^ = m. 

A3 

In Fig. 163, BCDE is a rectangle having a constant altitude 
a and a variable base x. When a; takes an increment Ax, 
the area A will take an increment aAx. 



.\ AA = aAx 
When two variables are so related that the ratio of their 



AA 
or -r— = a. 
Ax 



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§189] ELEMENTS OF CALCULUS 243 

corresponding increments is variable, either variable is said to 
change non-uniformly with respect to the other. 

If the variables s and t are related by the equation s = \gt 2 } 
then As = \g(2tAt + At 2 ). See Art. 167, example 2. 

As 
Here ^r is a variable for it varies with t, that is, differ- 
As 
ent values of t give different values of tt, and the change is 

non-uniform. 

189. Differentials. — If two variables are so related that 
one is dependent and the other is independent, then for 
corresponding values of the variables: 

(1) The differential of the independent variable is the value 
of its increment. 

(2) . The differential of the dependent variable is what would 
be its increment, if at the corresponding values considered, 
its change became and remained uniform with respect to the 
independent variable. 

The differential of a variable is denoted by writing d before 
it. 

Thus, differential x is denoted by dx. Also dy, d(x*) t 
d(x 2 + 2x + 1), and df(x) denote the differentials of y, x* } 
x 2 + 2x + 1, and/(x), respectively. 

190. Illustrations. — It follows from the definitions that the 
differentials of variables that change uniformly with respect 
to each other, are their corresponding increments. 

Thus, if y = mx + b, dx = Ax and dy = Ay, for y changes 
uniformly with respect to x. 

It should be noted that dy = Ay when, and only when, the 
graph of y = f{x) is a straight line. 

If the rectangle of constant altitude, Fig. 163, is increased in 
area by increasing the base by the length CM , the area is 
-increased by the rectangle CMND. Here evidently the area, 



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244 ANALYTIC GEOMETRY [§190 

A, is a function of the base x. Since A and x change uniformly 
with respect to each other, CM = dx and the rectangle 
CMND - dA. 

Consider the curve y = /(x), Fig. 164, as being traced by 
a point starting from the origin and moving to the right and 
upward. The direction that the tracing point is moving at 

any point is along the tangent line 
at that point. 

Let (x, y) be the coordinates of 
the moving point. 

Evidently, y is changing non- 
uniformly with respect to x. 

Suppose the moving point has 

reached Pi. Here y is evidently 

Fig. 164. changing at the same rate it would 

if the point were moving along the 

tangent line at P x . If then the change in y is to become and 

remain uniform with respect to x, the point must move 

along the tangent. 

It follows that at the point Pi, if the increment of x is 
Ax = MiM, dx = Ax, and dy = QT. 

It is to be noted that the corresponding increment of y is 

Ay = <2P. 

dy 
Further, if the slope of the tangent, -r-, that is, the deriva- 
tive, is represented by/'(x), 

dy = f (x)dx. 
Since dy and dx are finite quantities, dividing by dx, 

This is an extremely important and useful relation, for it 
states that the derivative and the ratio of the differentials can 
be used interchangeably. 

Again, referring to Fig. 164, if « is the length of the curve 



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§190] ELEMENTS OF CALCULUS 245 

traced, then corresponding to dx and dy, the change in s, if 
this change becomes and remains uniform, is ds = PiT, and 

ds 2 = dx 2 + dy 2 . 

The triangle PiQT is called the differential triangle. 

Example. — A point is moving along the parabola y = 3x*. When it 
has reached the point whose abscissa is 2, find dy and ds corresponding 
to dx = 0.1. 

Solution. — First find the derivative of y with respect to x. 

Given equation y = 3x*. 

dx 
;. dy = 6s -dx, for any value of x. 

When x - 2 and dx * 0.1, d y = 6-2-0.1 - 1.2. 
And da - V<te* + dy 2 - \/0.1 2 + 1.2* - 1.2042—. 

EXERCISES 

1. The right triangle, Fig. 165, is being generated by the altitude 
moving uniformly to the right. If the variable base is x and the area 
A, show that dA corresponding to dx is the rectangle M\MQP\. 

2. The area of the upper half of the area of 
the parabola y* — 4x is being generated by the 
ordinate moving toward the right. If A is the 
variable area, show that dA — 2\/x dx for any 
value of x. Draw the figure. 

3. If the upper half of the area of the circle 
x s + y* = r* is being generated by the ordinate 
moving uniformly toward the right, show that 
dA =* y/r* — x* dx. 

4. The area above the s-axis between y « sin x and the s-axis is being 
generated by its ordinate. Show that dA = sin x dx. For the part 
below the z-axis show that dA = —sin x dx. 

5. A point is moving on the circle x* + y* = 25. Find dy and d* 
corresponding to a change in x of dx = 0.2 at the point in the first 
quadrant where x » 3. 

6. A point is moving on the ellipse 52 +ii ■ !• Find dy correspond- 
ing to dx = 0.4 at the point in the first quadrant where x — 2. In the 
second quadrant where 3 « — 2. 




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246 ANALYTIC GEOMETRY [§191 

In the following find dy for any x. 

7. y - 3s 1 + 2x - 1. 11. y = Vx* + 4. 

8. y = x 8 + 4x + 2. 12. x^ + ^ - 4. 

9. y » s« - 3x» + 2x*. 19. x* + ^ - 2. 

m* £_ L 14 . y ,1 



16 9 * VSm 1 ^ 

15. The distance 8 that a body will fall in t seconds is given by 
the formula * = \qt*. Find d« for any value of t. Find ds when t — 2 
and corresponding to (ft = 1. (Use g = 32.) 

INTEGRATION 

191. The inverse of differentiation. — Just as division is 
the operation that is the inverse of multiplication, and the 
extraction of a root is the inverse of raising to a power, so 
differentiation has its inverse operation. Here, as usual, the 
inverse operation is the more difficult. In fact, it is frequently- 
impossible to do the inverse of a differentiation except 
approximately. 

The process of doing the inverse of a differentiation is 
called integration. The result obtained is called an integral. 

The methods of integrating can be dealt with here to only a 
very limited extent. In general, an integral is found by 
knowledge acquired from differentiation, by reversing the 
rules of differentiation, or by reference to a table of integrals. 

Integration has very many applications to problems arising 
in the sciences and in engineering as well as to problems in 
mathematics. 

The symbol, ,/*, indicates that the differential before which 
it is written is to be integrated. 

Thus, J*2xdx indicates that a function of £ is to be found whose 
differential is 2xdx. The function is evidently z* + C % where C is any 
constant, for 

<*(*» + O - d(x \ + Q dx - 2xdx. 
ax 

Since the differential of any constant is zero, the function 

sought when integrating may contain a constant no indication 

of which appears in the given differential. For this reason 



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§192] 



ELEMENTS OF CALCULUS 



247 




the integral of a different'al is, in general, indefinite, and is 
called an indefinite integral. 

The constant C that is supplied when integrating is called 
the constant of integration. 

192. Determination of the constant of integration. — The 
constant of integration is determined by 
having some fact about the function 
given besides its differential. This can 
be best illustrated by examples. 

Example 1. — Find the equation of a curve 
such that the slope of its tangent line at any 
point shall equal to the abscissa of the point 
if, further, it is given that the curve passes 
through the point (2, 4). 



dy 
Solution. — Since — = slope of tangent, and 
dx 

x = abscissa of point of tangency, 

dy 

dx 

dy 
Considering — as the ratio of dy to dx, and multiplying by dx f 
dx 

dy = xdx 

Then J*dy = fxdx, and y = ix* + C. 

Here C is any constant, and the equation represents all parabolas 
having their axes on the y-axis and opening upward. Some of these are 
represented in Fig. 166. 

It is evident that one such parabola can pass through any particular 
point of the plane. The one sought passes through (2, 4), and there- 
fore these values must satisfy the equation y » ix* + C. 

Substituting (2, 4) in this equation, 

4 = J.2* + C. :. C - 2. 

The equation of the curve satisfying both conditions is then 
y - iz* + 2. 

Example 2. — Find the area enclosed by the parabola y* =* 4x and 
the double ordinate corresponding to x = 8. 

Solution. — The parabola y 2 = 4x is shown in Fig. 167, and is sym- 
metrical with respect to the x-axis. Then one-half of the area is above 
the x-axis and is the area OMC. 



x. 



Fio. 166. 



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248 



ANALYTIC GEOMETRY 



[5193 




H-I 



Consider the area A as generated by an ordinate moving from the 
origin toward the right. When it has advanced any distance x 

dA — y dx. 
But y = +2y/x since y is positive. 
Then dA - 2y/x dx. 

Integrating. A - fr* + C. (1) 

A further fact about the area A is that A — 0. when x « 

Substituting these values in (1) gives 

0-0 + C. .\ C - 0. 
Hence for any value of x t A « \x* + 0. 
Andforx=8, A - l-8*=yV2 =30.17-. 
. * . the total area = 2A — 00.34— square 
units. 

193. Methods of integrating. — 

While a knowledge of differentiation 
enables one to write at once the in- 
tegrals of many differentials, the 
following formulas will help in inte- 
Fio. 167. grating forms that occur frequently. 

(i) yvdu - iL-j + c 

Here u may be any function of which du is the differential, 
and n is not equal to — 1. 

That (1) is true can be readily proved by finding the differ- 

ential of — r— r + C. 
n -jr 1 

Example 1. — Find fx*dx. 

Here x = u, dx — du t and n — 4. 

. * . ys 4 ^ =» fr 1 + C. 

Example 2.— Find f(x* + x*)*(3x* + 2a0<k. 
Here x* + a 1 - w, (3a; 1 -f 2s)<fo — du, and n — 2. 

. * . y (*» + x*)*(3a; J + 2s)d s - ./Vdt* - Ju»+ C - *(*« + *»)• + C. 
Example 3.— Find jf Vs* - 1 2xdx. 
Here a* — 1 = w, 2a;da; = du, and n = J. 

.-. yVa^H 2a**r - yw*dw - |w* + C - §(*« - 1)1 + C. 

(2) /cdu = c/du. 



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§193] ELEMENTS OF CALCULUS 249 

This states that a constant can be written either before or 
' after a sign bf integration. 

Proof. — Since d(cu) = cdu. By differentiation. 
Then cu = J* cdu. By definition of an integral. 

But fdu = u. By (1) where n = 0. 

And cfdu = cu. Equating values of cu 

.'. J* cdu = cfdu. 
Example.— fbx*dx - bfx*dx - 6 • \x* + C - fa: 4 + C. 
(3) y (du + dv) - S&vl + f&v. 

Proof. — d(u + v) = du + dr. By differentiation. 
Then y*(dw + dv) = w + v. By definition of an integral. 
But u = fdu and t; = y\fo. By (1). 

. * . f (du + dv) = y*dw + fdv. 
This can readily be extended to the integral of the sum of 
any number of differentials. 

Example.— f(x* + 3s* - x + l)ds 

= fx*dx + y3aj*da; - yxcfo; + fdx By (3). 
-' \x* + a' - fcr 2 + a + C. By (1) and (2). 

Here C is the sum of the several constants of integration. 

EXERCISES 

Find the indefinite integrals in exercises 1-10, and check by differ- 
entiation. 

1. dy = 4xdx. 6. dy - (2x + l)dx. 

2. dy « x*dx. 7. dy = (2a; 2 + x + 2)da\ 

3. dy « 4a;*da;. 8. dy - (a; — l)(x + l)ds. 

4. dy - aAto. 9. dy - (a; + l) 8 da;. 

5. dy « a^ds. 10. dy = (a + l)*da\ 

11. Find the equation of the curve whose slope at any point is equal 
to three times the abscissa of that point, and which passes through the 
point (2, 6). 

12. Find the equation of the curve whose slope at any point is equal 
to the square of its abscissa at that point, and which passes through 
the point (1, 1). 

13. Find the equation of the curve whose slope at any point is equal 
to the square root of its abscissa at that point, and which passes through 
the point (2, 4). 



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250 



ANALYTIC GEOMETRY 



[§194 



14. Find the area enclosed by the parabola y 1 — 2x and the double 
ordinate corresponding to x * 4. 

15. Find the area enclosed by the parabola y* — 3x, the z-axis, and 
the ordinate corresponding to a: — 2 and x = 8. 

16. Find the area between the curve y — 2a& and the z-axis from 
the origin to the ordinate corresponding to x = 10. Check by finding 
the area considered as a triangle. 

17. Find the area between the curve y - x x and the z-axis from the 
origin to the ordinate corresponding to x '*» 4. 

18. Find the area between the curve y — x % and the z-axis from the 
ordinate corresponding to x = —3 to the origin. 

19. Find the area enclosed by the semi-cubical parabola y = z* and 
the double ordinate corresponding to x — 4. 

20. Find the area enclosed by the curve y — z""*, the z-axis, and 
the ordinates corresponding to x = J and z = 8. 

21. Find the area that is below the z-axis and is enclosed by the 
parabola y — z* — 4z + 3 and the z-axis. 

22. Find the area that is below the z-axis and is enclosed by the 
curve y = z» — 4z* + 3z and the z-axis. 

TRIGONOMETRIC FUNCTIONS 

194. So far in the calculus a study has been made of alge- 
braic functions only. The trigonometric functions will now 

be considered to a limited ex- 
tent. The sine and cosine will 
receive the chief attention, the 
formulas of the others will be 
given for completeness only. 

195. Derivatives of sin u and 
cos u. — Let be a unit circle 
generated by the point P (x, y) 
moving in the positive direction, 
Fig. 168. 
Let u be the measure of the 
angle XOP in radians, and let s be the measure of the arc 
XP in linear units. 

Then u = s, x = cos u } and y = sin u. 

Differentiating, du = ds, dx = d(cos u), and dy = d(sin u). 




Fig. 168. 



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§195] ELEMENTS OF CALCULUS 251 

In the differential triangle PQT, 

dx = PQ, dy = QT, and ds = PT. 

Also the angle at T through which the tangent has turned 

is equal to u. 

Now dy = ds* cost*, cosw and dy having the same sign. 

But dy = d(sin u) and ds = du. 

. *. d(sinw) = cos u du. 

Dividing by dx gives the derivative formula: 

ttttt d(sia u) du 

VIH. -~i — - = cos u -j— 

dx dx 

Also dx = —ds sin u, sin u and dx having opposite signs. 

But dx = d(cos u) and ds = du. 

.'. d(cosu) = — sin udu. 

Dividing by dx gives the derivative formula: 

_ d(cos n) . du 

K. _L_J = _ smu _. 

It is to be noted that the derivation of VIII and IX requires 
that the angle shall be in radians. 

Example 1. — Given y = sin (3x* + 4x — 1), find ~ 

Solution.- dg = d8in(3*»+4*-l) _ 

dx ax 

= cos (8** + te - 1) d(3 *' ^ " 1} - By VIII. 

h ,^+fc-D ii W) jW.fl t 

<ix dx dx ax 

^ - (6x + 4) cos (3x* + 4x - 1). 
ox 

Example 2. — Find the maximum and minimum points of the curve 

y — cos x. 

Solution. — -]r — — sin x. 
dx 

Putting -p = gives — sin x ■» 0. 

x = nx, where n = 0, ±1, ±2, • • •. 
For values of x near an even number of times x but less than x, — sin x 
is positive; and for values of x near an even number of times x, but 



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252 ANALYTIC GEOMETRY [§196 

greater than *-, — sin x is negative. Hence the curve is rising before and 
falling after x - 2nr. 
. * . maximum points are the points for which x =• 2nx. 
likewise minimum points are the points for which x — (2n + l)x. 
Example 3. — Find the area enclosed by an arch of the curve y * sin x 
and the z-axis. 

Solution. — The curve y « sin x is shown 
in Fig. 169. The area sought extends from 
r^JT x * to x — r. 

Consider this area A as being generated 
Fio. 169. k y ^ ordinate moving toward the right. 

Then dA — yds - sin x dx. 
And y dA - y*sin x dx. 

. * . A — — cos x + C. By the inverse of differentiation. 
When x - 0, A =» 0. 

. • . « -cos + C, or C - 1. 
When a? « r, A = —cos x + 1 = 2 = number of square units in area. 

196. Derivatives of other trigonometric functions. — The 

following formulas are stated for completeness. Their deriva- 
tion is not difficult and may be performed as exercises. 

_ d(tan u) 5 du 

X. — *-j — - = sec 2 u -=-• 

dx dx 

_ d(cot u) , du 

XI. -*-= — - = -csc 2 u^— 

dx dx 

_. d(secu) . du 

XIL — —i — - = sec u tan u -=-• 
dx dx 

__ d(csc u) . du 

Xlil. — ^-j — - = —esc u cot u j— 

dx dx 

_ t7 d(vers u) . du 

XIV. — ^ = sin u -i— 

dx dx 



XV. 



XVI. 



du 
d(sin~ 1 u) _ dx 
dx "~ VI - u 2 
du 
dCcos- 1 u) dx 



dx Vl - u 2 

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§197] ELEMENTS OF CALCULUS 253 



xvn. 
xvra. 

XIX. 

xx. 

XXI. 



du 
dCtaii- 1 u) dx 



dx 1 + u* 

du 

d(cot-* u) dx 



dx 1 + u* 

du 

d(sec- x u) _ dx 

dx " uVu 2 - 1* 
du 

dfosc-^u) _ ^ dx 

du 

d(vers _1 u) dx 



dx \/2u — u 2 

197. y*sin u du and y*cos u du. — Many integrals involv- 
ing trigonometric functions occur in the applications of cal- 
culus, but here attention will be confined to y*sin u du and 
y*cos u du. 

y* sin u du = — cos u + C. 

This is readily proved, for 
d(— cos u + C) = d( — cos u) = sin u du. 
y*cos u du = sin u + C 

For d(sin u + C) = d(sin u) = cos udu. 

Example. — Find J*Bin(2x + l)dx. 

If 2x + 1 - u, du = 2 dx. 

Then write ./"sin (2x + l)dx in the form lj*8in(2x -f l)2dx and it is 
in the form of J* sin u du. 

.\/sin(2s + l)<& - ifBm(2x + l)2dx = -Jcos (2s + 1) +C 

EXERCISES 
In exercises 1-20 find the derivatives. 

1. y ■» sin 3a;. 5. y - sin 3s cos 2x. 

2. y - sin 2 x. 6. y - tan 3s. 
8. y - cos (2x H- 1). _ sin a? 



4. y - sin * cos x. • V m C08 x " 



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15. 


V 


« sin* x\/sec x. 


16. 


V 


= m cot n qx. 
1 — cos X 


17. 


V 


~ 1 + cos x' 


18. 


P 


- tan 30 + sec Z$. 


19. 


P 


- J taD s - tan + $. 


20. 


V 


= 3 sin x — 4 sin 1 x. 


=» a 


(1 


— cos 6), find dx and dy, then 



264 ANALYTIC GEOMETRY [§198 

8. y - tan 1 5x. 

9. y — 3 sin x. 

10. y - sin (x» + x*). 

11. y « sin (x* + Zx - 4). 

12. y - co s 2 (3s + 2). 
18. y - Vsin 3x. 
14. y = sin'x cos*x. 

21. Given x = a(0 — sin 0) and y 
by division find -p • 

22. Find the area enclosed by one arch of the curve y = cos x and the 
x-axis. 

28. Find the slope of the tangent to the curve y — sin x at the point 
where x =■ Jr. Where x — 2. 

24. Find the slope of the tangent to the cycloid x — a(0 — sin 0), 
y — a(\ — cos 6) at the point where 6 = ix. Where = *■. Where 
=0. 

25. Find the maximum and minimum points, and the points of inflec- 
tion of the curve y — sin x. 

Find the indefinite integrals in exercises 26-33. 

26. f sin 3x dx. 80. jT sin x cos x dx. 

27. y* sin (3x — l)dx. 81. y* sin 1 x cos x dx. 

28. f cos 4x dx. 82. y* cos* x sin x dx. 

29. y* cos (4x — 2)dx. 88. y* sin* x cos x dx. 

34. Find the equation of the curve passing through the point (x, 0), 
if the slope of the tangent at any point is equal to the cosine of the 
abscissa of that point. 

EXPONENTIAL AND LOGARITHMIC FUNCTIONS 

198. Derivative of log e u. — We will first find -?- when 

y = logeX.* 
Let P(x h t/i) be a point on the curve. 
Then j/i = logcXi or Xi = & l . 

Let x = Xi + Ax y and X\ + Ax = e vl+ **. 
Subtracting, Ax = & l+ *» - e vi = e vi (e* v - iy. 

*In log,x, e = 2.71828 • • •, the base of the natural system, of 
logarithms. .:..-■ 



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§190 ELEMENTS OF CALCULUS 265 

Dividing by Ay, — = • 1 10 — = Xi • 



Ay Ay Ay 

Or ^L « L A ^ 

when Ax = 0. 

or ^= i iim r Ay 1 

dx xi Ay = Le^ - 1J 

But it can be shown that A lu ? A [ — Ay , 1 - 1. 

Ay = Le** — 1 J 

Then, dropping the subscripts, •¥- = -, or dy =- dx> 

cfx X X 

Evidently, if 2/ = log«u, then dy = - dw. 
Dividing by <fc,g = i g- 

d(log«u) _ 1 du 
dx u dz 

199. Derivative of log A u. — Let a be any base. Since by 
a theorem of logarithms, log u = log«vlog a e. 

Then <*^ = <*0^i ogo e. ByV 

XXm. .-. *flg*>-Jgloi*. By XXII. 

If, in XXIII, a = 10, logi w expresses the common logarithm 
of u, and 

djlogiou) _ 1 dit , , 
dx ~~ u dx ' 

where ilf = logioe = 0.4343 — . 

200. Derivative of a tt and e tt .— Let y = a". 
Then log e y = u log«a. 



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256 ANALYTIC GEOMETRY [§201 

Taking the derivative of each side of this equation by 
XXII and V, 

1 dy du* 

~ dy du 7 

«»• •■ ^ -*£«** 

If o is put equal to e, and noting that log«$ = 1, XXIV 
becomes 

XXV. ^=e«£- 

dx dx 

201. Derivative of u\ — Let y = u v , where u and v are 
functions of x. 
Then log e j/ = v log«w. 

Taking the derivative of each side of this equation by XXII 
and IV, 

ldy^vdudv. 
y dx" u dx dx ** ' 
dy v du , dv , 

Tx = * uTx+ y te Xo *< u 

m v du . „ cfo . 

The application of formulas VII, XXIV, and XXVI should 
be carefully distinguished. Formula VII is used when a 
variable is affected by a constant exponent; XXIV is used 
when a constant is affected by a variable exponent; and XXVI 
is used when a variable is affected by a variable exponent. 

It is customary in calculus to omit the base when writing 
logarithms to the base e, and to express the base when it is 
not e. 

Thus, log 5 means log*5. 



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§202] ELEMENTS OP CALCULUS 257 

202. Illustrative Examples. 

Example 1. — Given y = log (X* + 3x), find -=-• 



dy __ 

dx ~~ 


1 d(x 
x» + 3x 


2 + 3x) f 
da? 




(9-r 






xt + Sx U* , «,. 


. <fy a 


2x +3 




" da? 


= x» + 3x 





By XXII. 
By III, VII, V, I. 



Example 2.— Given y - logi (l + 3s), find -p. 



Tx - r+E - s — log10 *' By XXIIL 

1 31og 10 e. By III, V, I. 



l + 3x 

. dy 3 

* • " dx = 1 + 3x 



logio«. 



Example 3.— Given y - e* ,+x , find & 

p_^*d(£+z) t xx 

- e* ,+ * (2x + 1). By III, VII, I. 



Example 4. — Given y = log sin*x, find ™ 

dy 1 d(sin'x) 
dx sin*x dx 



By XXII. 



- gjig 2 sin x cos x. By VII, VIII, I. 

.'.^ = 2cotx. 
dx 

Example 5. — Given the catenary y = \a{eP + e a ), (see exercise 8, 
page 167), find the slope of the curve at the point whose abscissa is 0. 
17 



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258 



ANALYTIC- GEOMETRY 



[§203 



Solution.— Given y - \a{e a + e °). 



By HI, XXV. 
By V, I. 



. dy 
• 'dx 
dy\ 



J(e° - e °). 



ax\x = U 



203. 



the slope of the catenary at the point where x = is 0. 
J. I — f yVdu, and yVdu. — These integrals are readily 
evaluated and occur frequently. 

u 

For d(log.u + C) = — • By XXII. 

u 



£ 



= lOgeU + C. 




++»x 



Fig. 170. 

y*e tt du = e u + C. 
For d(e u + C) = e tt du. 

/aMu - t -^- + C. 



For d(r^— + C) = a tt du. 
Mog a / 



log a 



By XXV. 
By XXIV. 



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§203) ELEMENTS OF CALCULUS 259 

Example. — Find the area bounded by the equilateral hyperbola 
xy = 4, the x-axis, and the ordinates corresponding to x = 1 and x = 8. 

Solution. — The hyperbola is shown in Fig. 170, and the area sought 
is MNQR. 

Consider the area A as generated by the ordinate moving toward 
the right. 

Then dA = ydx. 

And fdA - fydx - f*dx - ±§~ 

A - 4 log x + C. 
When re - 1, A - 0. . ' . - 4 log 1 + C, or C - -4 log 1. 
When x = 8, A = 4 log 8 - 4 log 1. 

A - 4 X 2.079 -4X0 = 8.316. 
Therefore the area sought is 8.316 square units. 

EXERCISES 

In exercises 1-20 find the derivatives of the dependent variables 
with respect to the independent variables, and the differentials of the 
dependent variables. 

1. y = log (x 2 + 7x). 7. y = e 8 **". 

2. y = logio x*. 8. y = e* sin x. 

3. y = log -• 9. y = a 2 *. 

X 

4. y = logio or*. 10. y = xl0 2 * + *. 

5. y - e 2 *. 11. y - (3s - 2)*. 

6. y = e**. 12. y - J(« x + e"*). 

13. i = 6e-°«. (See Ex. 9, page 167.) 

14. i = /e L. (See Ex. 11, page 167.) 

15. y — e~* sin x. (See Ex. 18, page 174.) 

16. i - e~*' sin (2* + Jr). (See Ex. 19, page 174.) 

17. y = * + log (1 + x 2 ). 19. y = (3x + 2)<r* 2 . 

18. y - (2x + log x) 2 . 20. y = ( x * + 1) 2 *+ 8 . 

21. Find the slope of the tangent to the curve y = e* at the point 
where x = 0. Where x — 2. 

22. Find the slope of the tangent to the curve y — logio x at the point 
where x — 1. Where re = 10. 

23. Find the minimum point of the curve y = log (x 2 — 2x + 3). 

24. Find the maximum and minimum points of the curve whose equa- 
tion is y = 2x % — log x. 

25. Show that the rate of change of y with respect to x for any point 
on the curve y = oe* x is proportional to y. (See Art. 128.) 



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260 ANALYTIC GEOMETRY [§203 

26. Find the area bounded by the equilateral hyperbola xy — 1, the 
x-axis, and the ordinates corresponding to x - 1 and re — 10. 

27, Find the area bounded by the curve y ■* x + -> the x-axis, and 

X 

the ordinates corresponding - x — 2 and x - 4. 

Find the indefinite integrals in exercises 28 to 37. 



-dx. 34. y(l - *-i)(l - x"*)cix. 



SO. J^-tI«te. 85. y\»*<te. 

81 fco8*<te 

J SIT 



^±^<te. 



36. y(e* + 4)e-*d*. 

SUI X 

32, fe**dx. 37. y (e«* +1 + x)dx. 

38. Find the equation of the curve passing through the point (0, 1) 
if the slope of any point of the curve is proportional to the ordinate 
of the of that point. 

Suggestion.— ~r " ky. .'. — = kdx. 

ax y 

38. Find the equation of the curve passing through the point (0, 1) 
if the slope of any point of the curve is equal to xy. 

Suggestion.— -p — xy. .' . — = xdx. 

ox y 



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CHAPTER XIII 
SOLID ANALYTIC GEOMETRY 



204. Introduction. — In plane analytic geometry, all the 
points and lines are confined to one plane. In solid analytic 
geometry this restriction 'is removed, points and lines are 
considered as anywhere in space. In addition a new element 
is introduced, a surface of which the plane is a particular 
instance. 

Since plane analytic geometry is a special case of solid 
analytic geometry, it is expected that the formulas obtained 
for plane analytic geometry can be 
obtained as special cases of the 
formulas for solid analytic geometry. 
Such reductions and resemblances 
should be constantly sought. 

205. Rectangular coordinates in 
space. — If at the origin of the coor- 
dinate system in plane analytic 
geometry a line is erected perpendic- 
ular to the plane of the axes, this 
line will serve as a third axis for a 
space coordinate system, and is called 
the z-axis. It is customary in a space 
depiction to draw the x-axis, Fig. 171, horizontal, the 3-axis 
vertical and the #-axis as coming toward the observer. In 
order to give space perspective to the figure, the positive 
y-axis is drawn so as to make an angle of 135° with the posi- 
tive z-axis, and the unit on the j/-axis is taken equal to half 
the diagonal of a square whose side is a unit on the ic-axis. 

The three axes determine three coordinate planes, the 

261 



i 


z 




V 


B 


/ 


y 






O 






II 


M A 


/ 




/" 


? 




i 


£ 



Fig. 171. 



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262 ANALYTIC GEOMETRY [§205 

xy-plane, zz-plane and yz-plane. These three coordinate 
planes are mutually perpendicular to each other and all 
pass through the origin 0. 

If through a point P in space, Fig. 171, planes are drawn 
perpendicular to the x, y, and z-axes, respectively, these three 
planes will form with the coordinate planes a rectangular 
parallelepiped. The edge RP perpendicular to the yz-plane 
and parallel to the x-axis is called the x coordinate of P. It is 
considered positive if measured to the right, and negative if 
measured to the left. 

The edge NP perpendicular to the xz-plane and parallel 
to the y-axis is called the y coordinate of P. It is considered 
positive if measured toward the observer, and negative if 
measured away from the observer. 

The edge KP perpendicular to the xy-plane and parallel 
to the z-axis is called the z coordinate of P. It is considered 
positive if measured upward and negative if measured 
downward. 

These three coordinate lines uniquely determine a point P, 
since they determine three mutually perpendicular planes, 
MP, HP 9 and LP which intersect in one point P. 

In place of drawing a rectangular parallelepiped to rep- 
resent a point in space it is customary to draw a broken line 
consisting of three of its edges. 

Thus, the point P, Fig. 171, would be represented by the broken line 
OHKP. 

The three coordinates of a point are written (x, y, z.) 

Thus if P, Fig. 171, is the point (2, 3, 4), its coordinates are 
x - OH - 2, y = HK - 3, and z = KP - 4. 

The three coordinate planes divide all space into eight 
octants. The octant in which the point lies is denoted by 
the sequence of signs for the three coordinates. 

Thus, the (+, -f-, — ) octant is the octant to the right of the 
2/z-plane, in front of the sa-plane, and below the sy-plane. 



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$206] SOLID ANALYTIC GEOMETRY 263 

EXERCISES 

1. If P, in Fig. 171, has the coordinates (2, 3, 4) what are the coordi- 
nates of H, K, M, R, L and JV? 

2. Plot the points (1, 1, 1), (-1, 2, 3), (2, -3, 1), (-2, -1, -3). 

3. Draw the triangle whose vertices have the coordinates (2, 1, 4), 
(-1,3, 2), (2, -1, -3). 

4. Where are all the points for which x = 0? y » 0? z = 0? 

5. Where are all the points for which x = -2? y - 3? z - -2? 

6. From the point (zi, y\, Zi), perpendiculars are drawn to the codrdir 
nate planes. Find the coordinates of the feet of these perpendiculars. 

206. Geometrical methods of finding the coordinates of a 
point in space. — Since any point P in space c^tn be regarded 
as the vertex of a rectangular parallelepiped which has the 
opposite vertex at the origin, the codrdinates of P can be 
found geometrically in a number of different ways of which 
the following are the most useful. 

(1) From P draw a line, Fig. 171, perpendicular to the 
try-plane and meeting it in K. From K draw a line perpendic- 
ular to the x-axis and meeting it in H . 

Then OH - x, HK -•», and KP = z. 

(2) From P draw planes perpendicular to the x, y, and 
z-axes, respectively, and let the axes intersect these planes in 
the points H , M , and L, respectively. 

Then OH = x, OM = y, and OL = z. 

(3) From P draw lines perpendicular to the x, y y and z-axes 
meeting them in the points H } M, and L respectively. 

Then OH = x, OM = y, and OL = z. 

207. Distance between two points. — The distance between 
two points Pi(xi, j/i, Zi) and P 2 (x 2 , 2/2, Z2) is found by con- 
structing a rectangular parallelepiped, Fig. 172, having as 
opposite vertices Pi and P 2 , and whose edges are parallel to 
the coordinate axes. 

Then PiP 2 is the length of a diagonal of this rectangular 
parallelepiped. 



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264 



ANALYTIC GEOMETRY 



[§208 



Since PiSP* is a right triangle, P l P i t » P^ + SP S *. 
Since PiRS is a right triangle, P^S* - P\R* + RS*. 



Therefore 



iW - PiR* + R& + SPf. 



Substituting PiP» — d, Pi# = a* — Xi, AS — y* — yi, and 
SPi = z% — zi y and extracting the square root of both sides 
of the equation, gives the distance formula 

[«] d - V(xi - x*) 1 + (yi - y*)* + (ii - z,) 1 . 



J 


f 






p. 






/ 


s 


/ 






Pi 


^ 


s^ 




g 




^ 


r 




o 








y 










/* 


/ 


/ 


/ 






*x 



Fio. 172. 



Fig. 173. 



208. CoSrdinates of a point dividing a line segment in 

the ratio ri to r*. — As in plane analytic geometry, the ratio 

r P P 

— = n^ will be considered positive for internal division 

ft roTt 

and negative for external division. Let Pi and Pi, Fig. 173, 
be the end points of the segment and let P be the point of 
division. Through P it P 2 , and Po draw planes perpendicular 
to the s-axis meeting it in the points iVi, 2Vj, and iVo respectively. 
By a familiar theorem in solid geometry, 

PiPo _ NiNo 
PoPt ~ NoNi 

But NiNo = ON - ONi,ON = x©, and ONi = x h Art. 206. 
Hence iViiVo = xo — Xi. 

Similarly iVoiV2 = x% — xo. 



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§208] SOLID ANALYTIC GEOMETRY 265 

P P r 

Substituting these values and replacing p 1 ^ by -\ gives 





T\ Xo — * Xi 




Tt X2 — Xo 


Solving for xo, 






Xt\ = ■ 



n + r t 

By drawing planes perpendicular to the y-axis and the 
z-axis similar formulas are obtained for y and Zq. 

Therefore the coordinates of Po dividing the line P\Pi in 
the ratio r\ to r 2 are 

taoi ~ riXt + rgXi riy a + r 2 yi w riZ 2 + r^zi 

1491 Xo = r l + r 2 ' y ° = r,+r 2 ' Z ° = r> + u ' 

EXERCISES 

1. Find the distance between the points ( — 1, 3, 7) and (1, 9, 16). 

2. Find the distance between the points (3, —2, 4) and (6, —6, —8). 

3. Show that the points (1, -2, 3), (7, 0, 6), (4, 6, 8) form a right 
triangle. 

4. Show that the points (3, -1, 4), (4, 1, 7), (1, 4, 6) form a right 
triangle. 

5. Show that the points (1, 7, 6), (2, 2, 11), (2, 8, 13) form an isosceles 
triangle. 

6. Show that the points (-4, 2, 5), (-1, 5, 2), (-3, 3, 0) form an 
isosceles triangle. 

7. Show that the points (7, -1, 2), (4, 2, 2), (4, -1, 5), (3, -2, 1) 
are the vertices of a regular tetrahedron. 

8. Find the lengths of the medians of the triangle whose vertices are 
(-1.7,4)T, (3, -5, -2), (-5, 1,6). 

9. Find the codrdinates of the point which divides the line joining 
(7, 2, 6) to ( -3, 7, -9) in the ratio of 2 :3. 

10. Find the codrdinates of the point which divides the line joining 
(7, -6, 2) to (-3, 4, -5) in the ratio -3 :4. 

11. In what ratio is the line joining (2, —6, 3) to (4, —3, —6) divided 
by the xz-plane? 

12. The beginning of the line segment which is divided in the ratio 
4:3 by the point (1, 2, -6) is the point (-1, 6, -2). Find the 
codrdinates of the other extremity. 



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266 ANALYTIC GEOMETRY [§209 

18. Prove analytically that the straight lines joining the mid-points 
of the "opposite edges of a tetrahedron pass through a common point and 
are bisected by it. 

14. Prove analytically that the straight lines joining the mid-points 
of the opposite sides of any space quadrilateral pass through a common 
point, and are bisected by it. 

209. Orthogonal projections of line segments. — In general 
two lines in space will not intersect.. If parallels to these lines 
are drawn through any point, the angle made by these inter- 
secting lines is denned as the angle 
made by the non-intersecting lines. 

If through a point P in space a 
plane is constructed perpendicular to 
a given line, the point P* where the 
plane meets the line is defined as the 
orthogonal projection of the point P 
on the line. 
F 174 If the orthogonal projection of the 

end points Pi and P 2 of a line seg- 
ment, Fig. 174, on a line I are iY and P 2 ', then the line 
segment Pi'P 2 ' is said to be the orthogonal projection of the 
line segment PiP 2 on I. 

With these definitions it is easy to derive formulas for the 
projection of a line segment on a given line. 

Let PiP 2 , Fig. 174, be a line segment of length d, let Pi'P 2 ' 
be its projection on the line I, and let 6 be the angle between 
PiP 2 and P/P,'. 

Through P x draw a line parallel to I and meeting the plane 
passing through P 2 , perpendicular to line I in the point P 2 ". 
Join PtPt", P 2 "P 2 ', and PiPi'. Then is the angle P 2 "PiP 2 , 
and 

Pi'P 2 ' = PiP 2 " = PiP 2 cos 6 = d cos 0. 
This gives: 

Theorem 1. — The projection of a directed line segment on 
a given line is equal to its length multiplied by the cosine of the 
angle between the lines. 



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§210] SOLID ANALYTIC GEOMETRY 267 

Another theorem which is useful in solid analytic geometry 
is the following: 

Theorem 2. — The projection on any line of *the straight 
line joining any two points is equal to the algebraic sum of 
the projection of any broken line z 

joining these points. 

Proof.— Let P x P 2t Fig. 175, be 
the' straight line joining Pi and 
Pa, let I be the line on which 
PiP* is to be projected, and let 
P1P3P4P6P2 be any broken line 
joining Pi to P*. If the points 
P/, P 8 ', P/, P 6 ', P* are the pro- 
jections of the points Pi, P 3 , Pa,y 
Pby Pi, respectively, then Fio. 175. 

Proj. P1P3 + proj. PsP* + proj. P4P5 + proj. P^P* = 
Pi'Pa' + Pz'Pl + Pt'Pt' + Pi'P,' = Pi'P*'. 
But proj. PiP 2 = PiTY. 

This proves the theorem. 

210. Direction cosines of a line. — Let the angles which 

any line in space makes with the 
positive x, y, and 2-axes be re- 
spectively, a, P, and 7. These 
angles are called the direction 
angles of the line. Their cosines, 
cos a , cos ($, cos 7 are called the 
" wr direction cosines of the line. 

If Pi and P 2 , Fig. 176, are any 
two points on a line and d is the 
distance P1P2, the direction cosines 
are given by the formulas 

[60] cos « = ? 2 -=^» cos = ?*-=^> cos y - -*-^— • 




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268 ANALYTIC GEOMETRY [§210 

To prove this, let the projection of P\P* on the z-axis be 
Px'Pi', then 

Pi'Pt' = dcosa. 

But Pi'P,' = OP t ' - OPS and OP*' = a* and OPS = x x . 
Art 206. 
Therefore 

Xi — Xi = d cos a, 

or x 2 — x\ 

cos a = — 3 

a 

The remaining two formulas are found by projecting PiP* 
on the y-axis and the z-axis respectively. 

These three direction cosines are not independent, for squar- 
ing each equation and adding gives 

cos 2 a + cos 2 + cos 2 ? = 

(a* - xQ 2 + (y 2 - y x y + (z t - z x y d* 

d* d* ' 

Therefore 

[61] COS 2 a + COS 2 + COS 2 ? = 1. 

Example. — Find the direction cosines of a line if they are proportional 
to the numbers 2, —9, 6. 

Solution. — Since cob a : cos p : cos y —2: —9:6, 

cos a = 2k, 
cos = — 9k, 
cos y = 6fc. 

But by [51], the sum of the squares of these cosines equals unity, 
therefore 

4fc» + 81fc» + 36fc* - 1. 



Or 









fc= ± 


IT 








Substituting, 




cos a 


2 


cos/3 = 


9 


cos y — 


6 
11 


r 


cos a = 


2 


cos/3 = 


9 

= yz > COS 


y s — 


6 
11* 














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rfe 



§211] 



SOLID ANALYTIC GEOMETRY 



269 



EXERCISES 

Find the direction cosines of the lines joining the points in exercise 
1-3, and the projections of these lines on the three axes. 

1. (4, 2, 3) to (5, 3, 4). 3. (-5, 1, 4) to (-3, 4, -2). 

2. (-2, 1, 7) to (5, -3, 2). 

Find the direction cosines which are proportional to the numbers in 
exercises 4-7. 

4. -6, 2, -3. 6. 4, 3, -12. 

5. 6, -7, 6. 7. -10, -6, 15. 

8. Find the orthogonal projection of the line joining (7, 6, —2) to 
(5, —3, 4) on the x-axis; on the y-axis; on the z-axis. 

9. What are the direction cosines of the x-axis? Of the y-axis? Of the 
2-axis? 

10. What are the direction cosines of a line parallel to the z-axis? 
Perpendicular to the a?-axis7 

11. Where do all the lines lie for which (a) cos a = J, (b) cos = }, 
(c) cos a = i and cos = i, (d) cos a — 0, (e) cos a = 1? 

12. A line makes an angle of 60° with both the x and the y-axis, what 
angle does it make with the 2-axis? 

13. A line makes an angle of 75° with the s-axis, and 45° with the 
y-axis, what angle does it make with the z-axis? 

14. The equal acute angles which a line makes with the x-axis and 
the y-axis, are each one-half the angle which it makes with the z-axis. 
Find the direction cosines of the line. 

15. The angles not greater than 90° which ^ 
a line makes with the x, y, and z-axes are 
proportional to 1, 2, and 3. Find the direc- 
tion cosines of the line. \ *s 



N* 




211. Polar coordinates of a point. 
—If the distance OP, Fig. 177, of a 
point P from the origin is called p, 
and if the direction angles of OP are 
a, 0, and 7, then (p, a, 0, 7) are called 
the polar coordinates of P. The re- 
lations between the polar coordinates of P and its rectangular 
coordinates are obtained by replacing fa, j/i, Zi) of article 210 
by (0, 0, 0) and (x 2 , y 2 , z 2 ) by (z, y,z). 



Fig. 177. 



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270 



ANALYTIC GEOMETRY 



Since d « Vz* + V* + «S formula [60] gives the 
coordinates in terms of the rectangular coordinates. 

X 



11212 

polar 



cosa 



[52] 



±\/x* + y* + z* 

COS J = 7 y =$ 

±Vx* + y* + z* 



COS y = 



±Vx* + y 2 + z 2 
Note that the radicals must be taken either all positive or all 

negative. 

Replacing y/x 2 + y 2 + z 2 by its value p, and clearing of 
fractions gives the rectangular coordinates in terms of the 
polar coordinates. 

x = q cos a, 
[53] y - 9 cos ff, 

z = g cos y. 
Note that the direction cosines are not independent but are 
connected by the equation cos 2 a + cos 2 /8 + cos 2 ? = 1. 
212. Spherical coordinates. — Another method of locating a 
point in space is by means of spherical 
coordinates. From P, Fig. 178, drop 
a line perpendicular to the xy-plane 
meeting it in M. Join 0, called the 
pole, to P, and to M . Then the 
spherical coordinates of P are p, 0, 
and <f>, which are written (p, 6, <!>), 
where p = OP is the distance of P 
from the origin; 6 = angle NOM is 
the angle through which the positive 
x-axis would have to rotate to coincide with OM; and 
^ = angle ZOP is the angle which OP makes with the positive 
&-axis. 

The quantity p is taken positive if measured along the radius 




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§213] 



SOLID ANALYTIC GEOMETRY 



271 



vector, and negative if measured along the radius vector 
produced through the origin. The angle 6 can have any 
value from 0° to 360°. The angle 4> is restricted to values 
from 0° to 180°. 

The relations between spherical and rectangular coordinates 
are then 

z = q sin <p cos 6, 
y = p sin y? sin 0, 
z = q cos <p 

9 '= + Vx 2 + y 2 +z 2 , 

,-i I 



[64] 



tan 



y 

- 1 *- = sin~ 

X 



±Vx 2 + y 2 



^= COS" 1 



±Vx 2 + y 2 + z 2 



The convention with regard to signs is that either all the 
upper signs must be used, or else all the lower. 

If the pole of a spherical coordinate system were taken 
at the center of the earth, the z-axis passing through the 
north pole, and the xz-plane passing through the meridian of 
Greenwich, then the spherical coordinates of a point in the 
northern hemisphere can be so 
chosen that p will give the distance 
of the point from the center of the 
earth, its longitude and <p its co- 
latitude. 

218. Angle between two lines. — 
Let the two lines be h and U Fig. 
179, with direction angles a h fa, 71, 
and c*2, &, 72> respectively, and let 
$ be the angle between h and 1%. 

In order to find 0, draw two lines OP\ and OP 2 through 
parallel to h and k respectively, also draw ON, NM, MP h 
the coordinates of Pi, and let OPi = pi. 

By article 209, the angle between OP\ and OP 2 equals 6. 




Fig 179. 



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272 ANALYTIC GEOMETRY [8213 

Project OPi and the broken line ONMP x on 0P t . 
By theorem 2, Art 209, 

proj. OPi = proj. ON + proj. NM + proj. MPi. 
By theorem 1, Art 209, 

proj. OP i on OP i = pi cos 0, 
proj. ON on OP 2 = %i cos as, 
proj. NM on OP2 = j/i cos ft, 
proj. JfPi on OP 2 = 21 cos 72. 

Therefore picos = xicos at + yicos ft + 3icos 72. 

Replacing x\ f j/i, Si by their equivalents, [53], and dividing 
both sides of the equation by pi, gives the required expression 
for 0, 
[55] cos = cos ai cos a 2 + cos (3i cos § 2 + cos y t cos y 3 . 

If the two lines are perpendicular to each other, 

cos ai cos at + cos ft cos ft + cos 71 cos 72 = 0. 

If the two lines are parallel to each other, it is evident that 
either a\ = at, ft = ft, and 71 = 72, or ai = 180° — at, 
ft = 180° - ft, and 71 = 180° - 72. 

Example 1. — Find the polar coordinates of the point (1, — 1,— V2)- 
Prom [52], p — y/i — 2, cos a — i, cos — — i, cos 7 = — i\/2 

.-.« . 60°, - 120°, 7 - 135°. 
Then the polar coordinates of (1, -1, — \/2) are (2, 60°, 120°, 135°). 
If the negative sign is taken with p the polar coordinates are 
(-2, 120°, 60°, 45°). 
Example 2. — Find the spherical coordinates of (1, — 1, — V2)- 
From [64], p = Vi « 2, 

- tan" 1 (- 1) - sin- 1 3= * 315 °> **<*«> = cos- 1 -^-^? - 135°. 

Then the spherical coordinates of (1, -1, - y/2) are (2, 315°, 135°). 

If the negative sign is taken with each of the radicals the spherical 
coordinates are (-2, 135°, 45°). 

Example 3. — Find the direction cosines of a line which is perpendicular 
to two lines having direction cosines proportional to —1, 2, 6 and 
1, 4, 3 respectively. 



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§213] SOLID ANALYTIC GEOMETRY 273 

If cos a, cos 0, cos y are the required direction cosines, [55] and [51] 
give the three equations: 

—cos a + 2 cos + 6 cos y = 0, 
cos a + 4 cos + 3 cos y =0, 
cos* a + cos* + cos* 7 = 1.. 

Solving these equations, gives cos a'= f, cos = — f, cos y =f, or 
cos a — — $ , cos = $ , cos y = — f . 

Example 4. — Find the projection of the line segment U joining the 
points ( — 1, 3, 6) and (3, 7, — 1) on the line h joining the points (3, 1, —2) 
and (6, 7, 0). 

From [50] the direction cosines of h are $, J, — J and of U are 

f, I ?. 

If 6 is the angle between the two lines, by [55], 

12 + 24 - 14 22 



cos $ 



,63 63 



The length of h by [48] is d = 9, and the projection of U on U by 

9 X 22 22 
theorem 1, Art 209, is equal to d cos = — ~« — = -=r- 



EXERCISES 

Find the polar coordinates of the points in exercises 1-3, if their 
rectangular coordinates are: 

I. (1, V2, -1). 2. (4, -4, 4V2). 3. (1, 1, 1). 

4. If the polar coordinates of a point are (3, 60°, 60°, y) t find y. 
Find the spherical coordinates of the points in exercises 5-7, if 

their rectangular coordinates are: 

5. (2, 2V3, 4V3). 6. (-3, -\/3, -2). 7. (-\/6, VS, 2). 

8. Find the acute angle between the two lines having direction cosines 
proportional to 11, —10, 2 and —5, 2, 14. 

9. Find the direction cosines of a line which is perpendicular to two 
lines having direction cosines proportional to 2, 4, —3 and —1, 4, 3, 
respectively. 

10. Find the projection of the line segment joining (3, —1, 4) to 
(4, 1, 6) on'the line joining (4, 2, -5) to (-2, 4, -2). 

II. Find the projection of the line segment joining (7, 2, —3) to 
(2, 4, 3) on the line joining (1, -4, 3) to (7, -11, -3). 

12. Find the projection of the line segment joining (2, 1, —3) to 
(-2, 3, 1) on the line joining (3, -10, 4) to (12, 8, -2). 

13. Verify the conventions used with regard to signs in article 212. 

18 



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274 ANALYTIC GEOMETRY [§214 

SURFACES 

214. Locus in space. — If an equation in two variables is 
given in plane analytic geometry, values can be assigned at 
pleasure to one of these variables, and then the other is 
determined. The locus of all points satisfying such an 
equation is found in general to be a curve. 

On the other hand, in solid analytic geometry, if an equation 
in three variables is given, values can be assigned at pleasure 
to two of the variables and then the third variable is deter- 
mined. For instance, in the equation z — x 2 + y 2 , to every 
pair of values of x and y there corresponds a value of z. Hence 
for every point in the xy-plane there will be a corresponding 
point in space for the locus of z = x 2 + y 2 . If these points 
are thought of as a whole, it is obvious that they all lie on a 
surface. In general then the locus of a single equation in 
space is a surface. Sometimes one or even two variables may 
be missing in an equation, in which case such an equation 
will give rise to a special surface. 

215. Equations in one variable. Planes parallel to the 
axes. — The equation x = a, is satisfied by all values of y and 
z, since these variables can be regarded as entering into the 
equation x = a with zero coefficents. 

Hence all the points satisfying x = a will lie in a plane 
parallel to the yz-plane and cutting the x-axis at the point 
x = a. 

If the equation has the form f(x) = 0, the locus will consist 
of a series of planes, all parallel to the j/2-plane and cutting 
the x-axis at points whose abscissas are the roots of f(x) = 0. 

Like considerations hold for equations which contain only 
the coordinate y, or only the coordinate z. 

216. Equations in two variables. Cylindrical surfaces. — 
A cylindrical surface is generated by a straight line which 
moves so as to be always parallel to some fixed line, while 
intersecting a fixed curve. The fixed curve is called the 
directrix of the cylindrical surface, and the moving line in 



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§216] SOLID ANALYTIC GEOMETRY 275 

any one of its positions on the surface is called an element of 
the cylindrical surface. 

A plane can be regarded as a particular case of a cylindrical 
surface whose directrix is a straight line. 

Consider the equation x* + y 2 = 25, Fig. 180. In two 
dimensional space this is the equation of a circle with center 
at the origin and radius equal to 5. In 
three dimensional space, the coordinate 
z can be regarded as entering the equa- 
tion with a zero coefficient. Hence with 
any value of x and y which satisfies the 
equation, say x = 3 and y = 4, there can 
be associated any value of z. Thus, the . 
points (3, 4,-1), (3, 4, 0), (3, 4, 2), and, 
in general, (3,4,2) where z has any value, 
will all be points on the surface. These 
particular points all lie on a line per- FlG# 180, 

pendicular to the afy-plane and passing through the point 
3=3, y=4 in the sy-plane. 

In like manner through every point on the circle x 2 + y 2 = 25 
in the xy-plane there passes a line perpendicular to the xy- 
plane, and every point of this line satisfies the equation 
x 2 + y 2 = 25. Hence the locus of the equation x 2 + y 2 = 25 
is a cylindrical surface with elements perpendicular to the xy- 
plane, in other words parallel to the 2-axis, and having the 
circle x 2 + y 2 = 25 as directrix. 

Another illustration is the surface z 2 = x. Its elements are 
parallel to the y-axis and its directrix is the parabola z 2 = x 
in the a#-plane. This is called a parabolic cylindrical surface. 

In general an equation f(x, y) = represents in space a 
cylindrical surface whose elements are parallel to the 2-axis, 
and whose directrix is the curve f(x, y) = in the sy-plane. 

The equations f(y, z) = and /(x, z) = represent cylin- 
drical surfaces similarly situated with reference to the x and 
the y-axes, respectively. 



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276 



ANALYTIC GEOMETRY 



[§217 



217. Spheres. — Let C(h, k y I) be the center of a sphere of 
radius r. Since every point P on the sphere is at the constant 
distance r from its center 

CP = r. 

Or V(x - h) 2 + (y- k) 2 + (z - I) 2 - r. 

[66] .'. (x - h)« + (y - k)« + (z - 1)« = r*. 

This is the equation of the sphere with center at C and 
radius r. 

218. Surfaces of revolution. — A surface formed by revolv- 
ing a curve about a line in its plane is called a surface of 

revolution. The simplest 
cases are those where the 
curve is revolved about one 
of the coordinate axes. Sup- 
pose it is desired to revolve 
the parabola y 2 = x about the 
x-axis. Let P, Fig. 181, be 
any point on the parabola. 
As the curve revolves about 
the x-axis, P describes a circle of radius NP. When P is in 
the xy-plane, NP 2 = x. 

When P takes another position, say P', then NP' = NP and 
therefore NP' 2 = x, but NP' = \NM 2 + MF 2 = %* + z 2 . 
Replacing NP' by its value gives (y/y 2 + z 2 ) 2 = x, or 
y 2 + z 2 = x. 

Since P' can be any point on the surface, this is the equation 4 

of the surface of revolution. 

This equation was obtained by replacing y by y/y 2 + z 2 . 
If f(x : y) = is the equation of any curve in the xy-plane 
which is to be revolved about the x-axis, the same method of 
reasoning shows that /(x, \/y 2 + z 2 ) = is the equation of 
the surface of revolution. In like manner the equation of the 
surface of revolution obtained by revolving /(x, y) = about 
the y-axis is /(Vx 2 + z 2 , y) = 0. 




Fiq. 181. 



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§2181 SOLID ANALYTIC GEOMETRY 277 

Thus if y 2 = x is revolved about the y-ax is, the equation 
of the surface of revolution is y 2 = \/x 2 + z 2 , or y A = x 2 + z 2 . 

Similar formulas hold true if the curve is given in one of the 
other coordinate planes and revolved about the corresponding 
coordinate axes. For instance if the curve /(y, z) = is 
revolved about the g-ax is, the equation of the surface of 
revolution is f(\^x 2 + y 2 , z) = 0. 

If a circle and a line are in the same plane, and the line 
does not intersect the circle, the surface formed when the 
circle revolves about the line is called an anchor ring or torus. 

EXERCISES 

1. What is the equation of the plane parallel to the xy-plane and 3 
units above it? 4 units below it? 

2. What is the equation of the xy-plane? Of the xa-plane? Of the 
yz-plane? 

8. What is the equation of the locus of a point distant 3 units from 
the x-axis? 4 units from the z-axis? 

Find the equation of the locus of a point determined by the conditions 
in exercises 4-9. 

4. Equidistant from the points ( — 1, 2, 3) and (3, 4, —2). 

5. Equidistant from the xy-plane and the xz-plane. 

6. Equidistant from the z-axis and the y-axis. 

7. Equidistant from the x-axis and the yz-plane. 

8. Equidistant from the point (2, —4, 3) and the x-axis. 

9. The sum of the squares of its distances from the point (1, 1, 1) 
and (2, — 1, 3) is constant and equal to 17. 

10. Find the equation of a sphere with center on the x-axis, radius 
equal to 9, and which passes through the point (2, 4, —8). 

11. Find the equation of a sphere with center in the xy-plane, radius 
equal to 7, and which passes through the points (3, 4, 6) and (7, 3, 3). 

12. Find the equation of a sphere passing through the points (2, 3,-6), 
(5, 3, -5), (5, -2, 10), and (-3, 6, 6). 

Find the equations of the surfaces of revolution obtained by revolving 
the curves in exercises 13-24 about the axes as indicated. 
18. y = x, about the x-axis. 

14. y — s, about the y-suda. 

15. y a x 1 , about the x-axis. 

16. z* = x, about the x-axis. 

17. x* =■ -2«, about the x-axis. 



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278 ANALYTIC GEOMETRY [§219 

18. x* ■» 2z, about the z-axis. 

19. x* — 2« + z % — 0, about the x-axis. 

20. a;* — 2s + 2* — 0, about the z-axis. 

21. y =■ sin z s about the z-axis. 

22. y ■» sin «, about the y-axis. 

28. The ellipse - x + ?| » 1, about its major axis. This surface is 
called a prolate spheroid. 

24. The ellipse — t + j- t = 1, about its minor axis. This surface is 

called an oblate spheroid. 

25. What is the equation of the anchor ring obtained by revolving 
the circle lying in the ay-plane, with center at the point (0, 4) and 
radius 2, about the x-axis. 

Skeu;h and describe the following surfaces. 

26. x - y = 0. 81. a* + 4z l - 4. 

27. x* + y* - 4. 82. a* - 4z» - 4. 

28. x* - 2x + y« - 0. 88. a* + y* + z* - 2z - 2y - 2* - 6 = 0. 

29. y 2 - 2y + *« = 0. 34. a* - 3x + 2 - 0. 
80. y* - 2« - 0. 85. y* - 1 = 0. 

CURVES IN SPACE 

219. Equations of curves. — Since a single equation in three 
dimensional space is the equation of a surface, two equations 
will be satisfied simultaneously by all the points lying on the 
intersections of the two surfaces. In other words, it takes two 
equations in solid analytics to define a curve. 

Thus, y = is not sufficient to define the equation of the 
x-axis, for every point in the zz-plane satisfies this equation. 
Neither is z = sufficient, for this is satisfied by every point 
in the xy-plane, but y — and 2 = are satisfied only by 
those points common to the sz-plane and the sy-plane, namely, 
the z-axis. Therefore, y = and z = are the equations of 
the x-axis. 

Similarly x = and y = are the equations of the 2-axis; 
and x = and z = are the equations of the y-axis. 

Since it is evident geometrically that an unlimited number 
of surfaces can be passed through any curve, and that any 
two of these surfaces will be sufficient to define the curve, any 



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§220] 



SOLID ANALYTIC GEOMETRY 



279 



space curve can be represented by an unlimited number of 
pairs of equations. Thus a circle in space has for its pair of 
equations, the equations of any two spheres passing through 
it, or the equation of any sphere and the equation of the 
plane in which the circle lies. Even this does not exhaust all 
the possibilities of representing a circle, since any two surfaces 
passing through the circle will define it. 

220. Sections of a surface by planes parallel to the coSrdi- 
nate planes. — The curve in which a surface is cut by a coordi- 
nate plane is called the trace of the surface in the coordinate 
plane. Thus, the sphere x 2 + y 2 + z 2 = 25 and the plane 
2 = define a curve, the circle formed by the intersection of 
the sphere and the xy-plsme. If z is put equal to zero in the 
equation of the sphere, it becomes x 2 + y 2 = 25. This is the 
equation of the trace of the sphere in the xy-plane. 

By putting y = or x = the 
trace of the sphere in the sz-plane 
or the $/3-plane is obtained. 

Consider the two equations 
x 2 + y 2 + z 2 = 25 and z = 3. 

The curve AB, Fig. 182, com- ., % 

mon to these two surfaces is 
known from solid geometry to 
be a circle. If 3 = 3 is substi- 
tuted in x 2 + y 2 + z 2 = 25, it 
becomes x 2 + y 2 = 16. This 
is the equation of a circular cylinder. 1 Since every point 
satisfying the equation of the sphere and the plane satisfies 
the equation of the cylinder, this cylinder must pass through 
the circle AB. Hence substituting z = 3 in the equation of 
the sphere x 2 + y 2 + z 2 = 25, gives the equation of a cjdinder 
passing through the intersection of the plane 2 = 3 and the 
sphere x 2 + y 2 + z 2 = 25. 

1 For brevity the words cylindrical surface are often replaced by the 
word cylinder. 




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280 ANALYTIC GEOMETRY [§221 

The meaning of this substitution can be regarded from 
another standpoint. The equation x 2 + y 2 = 16 is the 
equation of the circle CD in which the cylinder cuts the 
ay-plane. All the elements of the cylinder are perpendicular 
to the £t/-plane and pass through the circle AB. Hence the 
circle CD is the projection of the circle AB in the xy-plane. 
In other words, substituting z = 3 in the equation of the 
sphere x 2 + y 2 + z 2 = 25, gives the equation of the projection 
on the xy-plane of the curve common to the plane z = 3 and 
the sphere x 2 + y 2 + z 2 = 25. 

In general, the substitution z = c, where c is some constant, 
in the equation of a surface can be regarded either as giving 
the equation of a cylinder passing through the intersection 
of z = c and the surface or as giving the equation of the 
projection on the #t/-plane, of the curve of intersection of the 
plane z = c and the surface. 

By giving c different values, the shape of different cross 
sections of the surface in planes parallel to the xy-plane are 
obtained. 

Like considerations hold for the substitution of x = a or 
y = b in the equation of a surface. 

221. Projections of curves on the coordinate planes. — 
When a curve is defined in space by two equations, it is 
desirable sometimes to know what are the equations of its 
projections in the three coordinate planes. 

Consider the curve defined by the equations 

x 2 + y 2 + z 2 = 49, (1) 

x 2 + 3y 2 - z 2 = 39. (2) 

If z is eliminated between these two equations, the resulting 
equation 

2x 2 + 4y 2 = 88, 
or x 2 + 2y 2 = 44 (3) 

represents an elliptical cylinder. Furthermore any point 
whose coordinates satisfy equation (1) and (2) will also 



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§221] SOLID ANALYTIC GEOMETRY 281 

satisfy equation (3), therefore the cylinder (3) passes through 
the curve defined by the equations (1) and (2). 

From another standpoint x 2 + 2y 2 = 44 is the equation of 
the directrix of the cylinder x 2 + 2y* = 44, and since the 
elements of this cylinder are perpendicular to the sy-plane, 
the equation x 2 + 2y 2 = 44 is the equation of the projection 
on the sy-plane of the curve defined by equations (1) and (2). 

In general, to find the equation of the projection on the xy- 
plane of the curve defined by the equation f\(x, y,z) = and 
f2(x y y, z) = 0, elminate z between these two equations. The 
resulting equation g(x, y) = is the equation of the projection 
on the xy-plane of the curve that is defined by the equations 
/i(z, y, z) = and f 2 (x, y, z) = 0. 

Proof. — Every point which satisfies simultaneously the 
equations fi(x } y } z) = and / 2 (x, y, z) = will also satisfy 
g(x, y)=0 and therefore g(x } y) = will pass through the inter- 
sections of these two surfaces. But g(x } y) = is the equation 
of a cylinder whose elements are perpendicular to the xy-plane. 
At the same time g(x f y) = is the equation of the trace 
of this cylinder in the ;ry-plane. Therefore, g(x y y) = is 
the equation of the projection on the xy-plane of the curve 
defined by the equations fi(x, y, z) = and/ 2 (x, y, z) = 0. 

In like manner it can be shown that to find the equation 
of the projection on the xz-plane of a curve defined by the 
equations fi(x, y, z) = and / 2 (x, y, z) = 0, eliminate y 
between these two equations; and to find the projection on 
the yz-plane eliminate x. 

For example, the projection on the x?/-plane of the curve defined by 
equations (l) and (2) is the ellipse x 2 + 2z % = 54, and projection on the 
2/z-plane is the equilateral hyperbola z % — y* = 5. 

EXERCISES 

Discuss and draw the traces on the three coordinate planes of the 
surfaces in exercises 1-3. 

1. x* + 2y* + 3s* = 6. 2.^+^+2-0. S. x* + y* - z - 1. 



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282 ANALYTIC GEOMETRY [§222 

Find the equations of the projections on each of the three coordinate 
planes of the curves in problems 4-9. 

4. x* - y* + z* - 4, 7. x* + y* - a*, 

2x» + y* - &* - 6. , s» + 2» - a*. 

5. x* - y - 0, 8. x* + y» - a*, 

2x* + y — * f ■■ 0. 2 = mx. 

6. x* + y* + 2* - a*, 9. y* + z* - 4as, 

** + y 1 — ox. y* = ox. 

10. Show that sections of - - t — 1^- = z are hyperbolas if perpendicular 
to the z-axis, but parabolas if perpendicular to the x-axis or y-axis. 

DISCUSSION OF EQUATIONS OF SURFACES 
222. Surfaces in space. — It is much more difficult to vis- 
ualize a surface in solid analytic geometry from its equation 
than to visualize a curve in plane analytic geometry from its 
equation. The following discussion similar to the one in the 
plane case is helpful. 

(1) Symmetry. 

(2) Intercepts on the axes. 

(3) Traces on the coordinate planes. 

(4) Sections of the surface by planes parallel to the coordi- 
naie planes. See Art. 220. 

(1) Symmetry. — To test the symmetry of a surface with 
respect to the coordinate planes, 

(a) replace x by —x, 

(6) replace y by -y, 

(c) replace z by —z. 

If the equation of the surface remains unchanged in case 
(a) it is symmetrical with respect to the j/z-plane, in case 
(6) with respect to the £2-plane, in case (c) with respect to 
the zy-plane. 

To test for symmetry with respect to the axes 

(a) replace y by — y and z by —z } 

(6) replace z by — z and x by — x, 

(c) replace x by — x and y by —y. 



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§222] SOLID ANALYTIC GEOMETRY 283 

If its equation remains unchanged in case (a) it is symmetri- 
cal with respect to the x-axis, in case (6) with respect to the 
t/-axis, in case (c) with respect to the z-axis. 

To test for symmetry with respect to the origin replace 
s by -x, y by -j/, z by -z. 

If its equation remains unchanged, ita surface is symmetrical 
with respect to the origin. 

(2) Intercepts on the axes. — To get the intercepts on the 
x-axis, set both y and z equal to zero in the equation of the 
surface and solve the resulting equation for x. The solutions 
of this equation are the intercepts on the x-axis. Similar 
considerations hold true for the t/-axis and the 3-axis. 

(3) Traces on the coordinate planes. — To get the trace of a 
surface in the xy-plane set z = 0. The resulting equation is 
the equation of the trace of the surface in the xy-plane. Similar 
considerations hold true for the traces in the sz-plane and 
the ys-plane. See Art. 220. 

Example. — Discuss and draw the locus of the 
equation x 2 + 2y 2 = z. 

(1) This surface is symmetrical to the ys-plane, 
the zz-plane and the s-axis. 

(2) Its intercepts on the three axes are 0. 

(3) Its traces are as follows: 
In the xy-plane, the point ellipse x 2 + 2y 2 = 0. 
In the xz-plane, the parabola x 2 — z. 
In the yz-plane, the parabola 2y 2 = z. 

(4) Taking sections by planes z = c shows the 
projections of these sections to be the ellipses 

?!.!£._ -I Fig. 183. 

c + ic ~ 1 ' 

If c<0, these ellipses are all imaginary, hence no part of the surface 
lies below the xy-plane. 

If c — 0, the equation x 2 + 2y 2 = 0, shows the section to be a point 
ellipse. 

As c increases from without limit, the ellipses increase in size without 
limit, the semimajor axis being \fc and the semiminor axis i\/2c, hence 
the surface is as pictured in Fig. 183. In this case it is not necessary 
to take sections parallel to the other coordinate planes. 




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284 



ANALYTIC GEOMETRY 



[§223 



QUADRIC SURFACES OR CONICOIDS 

223. General equation of second degree. — The locus of 
the general equation of the second degree, 

Ax 2 + By 1 + Cz* + Dxy + Eyz +Fxz+Gx+ Hy + Kz + L = 0, 
is called a quadric surface. It is also called a conicoid because 
every section of a quadric surface by a plane is a conic. By 
rotation and translation of axes it can be shown that this 
equation has for its real locus, five distinct types of surfaces 
besides cylinders, cones and degenerate forms like planes, 
lines and points. These five types will now be considered. 

224. Ellipsoid, g + g + g-i. 

(1) This surface, Fig. 184, 
is symmetrical to all the 
coordinate planes, all the 
coordinate axes, and the 
origin. 

(2) Its intercepts on the 
axes are x = ±a, y = ±b, 
z = ±c. 

(3) Its traces are as follows: 



i 


Z 


\^ X 




Y* 


\- r — J 



Fiq. 184. 



= 1. 



In the zy-plane, the ellipse 

In the xz-plane, the ellipse 

In the 2/z-plane, the ellipse j- 2 + - 2 = 1. 

o c 

(4) Sections of the ellipsoid by the planes z = k are the 
ellipses 

a 2"1- fe 2 l C 2, Z -*> 



X 

a 
a 



or 



+ 



b\ 



= 1, z = k. 



. (C 2 _• fc 2) - (C 2 _ J,) 



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§225] SOLID ANALYTIC GEOMETRY 285 

These ellipses have their centers on the s-axis, semimajor 
axes equal to -\c 2 — fc 2 , and semiminor axes equal to 

- Vc 2 — k 2 . Here it is assumed that a > b. If a < 6, the axes 
c 

are interchanged. 

As k increases numerically from to c, the axes decrease 
from a to 0, and from b to 0, respectively. When k is numeri- 
cally greater than c, the ellipses become imaginary. Hence 
the ellipsoid is contained between the planes z = — c and 
z = c. 

A similar discussion for the other axes shows that sections 
parallel to the other coordinate planes are ellipses, and that 
the ellipsoid is contained between the planes y = — 6 and 
y = b, and between the planes x = — a and x = a. 

The surface can be thought of as generated by a variable 

ellipse moving parallel to the zy-plane, with its center always on 

the s-axis, and the end points of its axes always on the ellipses 

x 1 z 2 , y 2 z 2 

-, + -,= l,and^ + - 2 = l. 

Special forms of the ellipsoid are the prolate spheroid when 
b = c and a>b, and the oblate spheroid when b = c and 
a<b. 

x 2 y 2 z 2 
225. The hyperboloid of one sheet — 2 + ^- 2 5 = !• 

(1) This surface, Fig. 185, is symmetrical to all the coordi- 
nate planes, all the coordinate axes and the origin, 

(2) Its intercepts on the axes are x = ±a, y = ±6. 

(3) Its traces are as follows: 

X 2 y 2 
In the a?2/-plane, the ellipse "2 + fr = 1 • 

x 2 z 2 
In the xz-plane, the hyperbola — 2 2 = 1. 

*2/ 2 z 2 

In the 2/s-plane, the hyperbola r^ 2 = 1. 



fe 2 



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286 



ANALYTIC GEOMETRY 



[§225 



(4) Sections of the surface by the planes z = k are the 
ellipses 

x 2 



* 2 + V* - i + fc2 



or 



+ 



y 2 



r,{c 2 + fc 2 ) 



b 2 

- 2 (c* + k 2 ) 



z •■*= k. 

= 1, z = k, 



These ellipses are real for all values of fc, increasing in 
magnitude as k increases numerically from to «. The 

smallest ellipse is the one for which 
k = 0, and this is the trace in the 
xy-plane. The intersections in 
planes parallel to the other axes 
are hyperbolas. 

This surface can be thought of 
as generated by a variable ellipse 
moving parallel to the xt/-plane, 
with its center on the z-axis, and 
the end points of its axes on the 
hyperbolas 

x 2 z 2 A y 2 z 2 

- 2 -- 2 =l,andp--- 2 = 1. 

The hyperboloid of one sheet 
Fig. 185. has the property that through 

every point on its surface there 
can be drawn two lines which lie wholly in the surface. The 
surface can be covered with a net work of two sets of lines. 
No two lines of the same set intersect each other, but any 
line of either set intersects every line of the other set. This 
surface can be generated by a line moving in such a way that 
it always intersects three other non-intersecting lines in space. 
It is called a ruled surface, because through every point on 
its surface, there can be drawn at least one line which lies 
wholly on the surface. 




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§226] 



SOLID ANALYTIC GEOMETRY 



287 



226. The hyperboloid of two sheets. 



- *- - - = 1. 

»2 K2 r 2 ** 



a* b 2 c* 

(1) This surface, Fig. 186, is symmetrical to all the coordi- 
nate planes, all the coordinate axes, and the origin. 

(2) Its intercepts on the a>axis are ±a. The intercepts on 
the y-axis and the z-axis are imaginary. 

(3) Its traces are as follows: 

/»»2 A*2 

In the a^-plane, the hyperbola — 2 — ~ = 1. 



In the zs-plane, the hyperbola — 2 ^ = 1. 

In the ys-plane, the imaginary ellipse ^ + -5 = — 1. 

c 




♦-X 



Fig. 186. 



(4) Although the trace in the ys-plane is imaginary the 
form of the equation suggests that sections parallel to the 
j/2-plane might be ellipses. Since it is easy to picture a surface 
in terms of increasing or decreasing ellipses, sections will be 
taken parallel to the #z-plane. Sections of this surface by such 
planes parallel to the j/z-plane as x = k, are the ellipses 



or 



2/ a 



S< fc2 -« 2 > 



t + z l = *_ 2 - 1 

ft* T C 2 a 2 l > 

- + *-* 



X = fc, 



# 2 - ^) 



-= 1, x = fc. 



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288 



ANALYTIC GEOMETRY 



[§227 



These ellipses are imaginary if —a<k<a. Hence there is 
no surface between the planes x = — a and x = a. As k 
increases numerically from a to », the ellipses increase 
indefinitely in magnitude. 

Sections by planes parallel to the other axes are hyperbolas. 

The surface can be thought of as generated by a variable 

ellipse moving parallel to the ys-plane, with its center always 

on the x-axis, and the end points of its axes on the hyperbolas 

x 1 y 2 , x 2 z 2 

7#- Si- ^ and ^-ri= 1. 



a 



a' 



227. Elliptic paraboloid. ^ + j£ =z. 

(1) This surface, Fig. 187, is symmetrical to the j/s-plane, 
the sz-plane, and the s-axis. 

(2) Its intercepts are x = 0, y = 0, and s = 0. 

(3) Its traces are as follows: 

X 2 y 2 
In the sy-plane, the point ellipse —^ + p = 0. 

In the zz-plane, the parabola x 2 = a 2 z. 
In the ys-plane, the parabola y 2 = b 2 z. 

(4) The trace in the sy-plane sug- 
gests that sections parallel to this plane 
might be ellipses, in fact, the sections 
of this surface by the planes, z = k are 
the ellipses 




~2 + P " fc > * - *■ 



+-X 



or 



x 2 y 2 
a 2 k ^ b 2 k 



= 1, z = fc. 



Fig. 187. 



If 



fc < 0, the ellipses have an im- 
aginary locus, hence no part of the 
surface lies below the xy-plane. As k increases from 
to a>, the ellipses increase in size indefinitely. The sur- 
face can be thought of as generated by a variable ellipse 



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§228] 



SOLID ANALYTIC GEOMETRY 



289 



moving parallel to the xy-plane whose center is on the 2-axis 
and the end points of whose major and minor axes are on the 
parabolas x 2 = a 2 z and y 2 — b 2 z. 

x 2 y 2 
228. Hyperbolic paraboloid. -, — ^ = z. 

(1) This surface, Fig. 188, is symmetrical to the yz-plane, 
the 33-plane, and the z-axis. 

(2) Its intercepts are x = 0, y = 0, and z = 0. 

(3) Its traces are as follows: 

x 2 y 2 
In the sy-plane, the two lines -| — j^ = 0. 

In the sz-plane, the parabola x 2 = a 2 «. 
In the yz-plane, the parabola y 2 = — b 2 z. 




Fig. 188. 

(4) Since no trace suggests an ellipse, and since it is easier to 
think in terms of moving parabolas instead of moving hyper- 
bolas, sections are taken by planes parallel to the yz-plane. 
Sections of the surface by the planes x = k are the parabolas 

These are parabolas, symmetrical to the xz-plane, opening 

(k 2 \ 
k, 0, -i) lying on the trace 

x 2 = a 2 z of the hyperbohc paraboloid in the xz-plane. All 
of these parabolas are congruent. Hence the hyperbolic 

10 



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290 



ANALYTIC GEOMETRY 



[§229 



paraboloid may be thought of as generated by a parabola 

opening downward of latus rectum 6 2 , moving with its vertex 

on x 2 = a 2 z so that its plane is always parallel to the t/z-plane. 

Sections parallel to the xz-plane are parabolas opening 

upward, and sections parallel to the rcy-plane are hyperbolas. 

This hyperbolic paraboloid has the property that through 

every point on its surface there can be drawn two lines which 

lie wholly in the surface. The surface can be covered with a 

network of two sets of lines. No two lines of the same set 

intersect each other, but any line of either set meets every 

z line of the other set. Hence the 

hyperbolic paraboloid is also a ruled 

surface. 

This surface can be generated by a 
line moving always parallel to a fixed 
plane, while always intersecting two 
"*" x non-intersecting lines in space. 




229. Cone. 



a 5 



+ U2 ~2 ~~ 0* 



(1) This surface, Fig. 189, is sym- 
metrical to the three coordinate 
planes, the three coordinate axes, 
Fig. 189. an( j ^he origin. 

(2) Its intercepts on the axes are x = 0, y = 0, and 2 = 0. 

(3) Its traces are as follows: 

x 2 v 2 
In the xy-plane, the point ellipse -, + ^ = 0. 

x 2 z 2 
In the xz-plane, the two lines —„ . = 0. 

* a 2 c 2 

v 2 z 2 
In the yz-plane, the two lines ^ 2 = 0. 

(4) Sections of the cone by the planes z = k are the ellipses 



x 2 y 2 
a 2 k 2 "*" b 2 k 2 



= 1. 



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§229] SOLID ANALYTIC GEOMETRY 291 

As the numerical value of k increases from to », the 
ellipses increase in magnitude indefinitely. Hence the surface 
can be thought of as generated by an ellipse moving parallel to 

x* z 2 
the xy-plane, with the ends of its axes in the lines -j * = 0, 

CL C 

XI 2 Z 2 

and the lines p , = 0. This cone is also a ruled surface, 

but it is covered by a single set only of lines, all of which pass 
through the origin. 



EXERCISES 

Discuss and draw the surfaces in exercises 1-15. 

8. x» + y* « 4«. 

9. x» - y* = 4z. 

10. x* + y* - z* « 0. 

11. xy H- xz H- yz — 0. 

12. cos a = 0. 

13. cos = £• 

14. cos = £• 

T *?"?~S" a 16. cos* = f 

Discuss and draw the curves or straight lines in exercises 16-20. 

16. x = 3, y = -2. 19. x = y = 2. 

17. cos a = cos = 0. 20. cos = £, cos </> *» J. 

18. y - x, x 2 + y* - 4. 

21. Find the equation of the locus of a point which moves so that 
the sum of the squares of its distances from the x and the y-axis equals 4. 
Discuss and draw the locus. 

22. A point moves so that the sum of the squares of its distances from 
two fixed points is constant. Prove the locus to be an ellipsoid. 

Suggestion. — Take the line through the two points to be the x-axis, 
and a point midway between them as the origin. 

23. A point moves so that the difference of its distances from two 
fixed points is constant. Prove the locus to be an hyperboloid. 

24. Find the locus of a point equidistant from the point (p, 0, 0) 
and the xz-piane. 



1. 

2. 


x» 




+ 4*» 
+ 4z» 


— 


9. 
9. 


3. 


X* 


-4y» 


- 4z* 


— 


9. 


4. 


X* 

9 


!/* z 2 
^ 4 ^16 


1. 




5. 


X 2 

9 


+*- 


z* 
"16 


1. 




6. 


X* 

9 


4 *" 


"16 


1. 





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292 



ANALYTIC GEOMETRY 



[§230 



THE PLANE IN SPACE 

230. Equation of a plane. — Some of the conditions that 
determine a plane in solid geometry are three points in the 
plane, or a point and a line in the plane. Unlike the straight 
line in two dimensional space, these simple conditions do not 
lend themselves readily to deriving the equation of a plane in 
three dimensional space. * Rather, one of the simplest ways of 
deriving the equation of a plane is by using the length of 
the perpendicular from the origin to the plane and the direc- 
tion cosines of this perpendicular. This perpendicular is called 
the normal to the plane. 

231. General equation of a plane. — Every equation of the 
first degree in x, y, and z as 

[57] Ax + By + Cz + D = 

represents a plane. 

Let P\{x\j yi, Zi) and P 2 (x 2 , y 2 , z 2 ) be any two points whose 
coordinates satisfy [57]. 

Then Axx + By x + Cz x + D = 0, (1) 

and Ax 2 + By 2 + Cz 2 + D = 0. (2) 

Take any two constants r x and r 2 , multiply equation (1) by 



rx + r 2 



multiply equation (2) by 



rx + r% 



, and add, 



A rxx 2 + r t xx , fi r x y% + r 2 y x c r x z 2 + r&x , D = Q 
rx + r 2 rx + r 2 rx + r 2 

This shows that any point on the line joining PxP 2 also 
satisfies equation [57]. Since Pi and P 2 are any two points 
on the surface [57], this shows that every line joining two 

* The equation of a plane through three points can be expressed in 
determinant form. If the three points are Pi, P s , and Ps, the equation is 

x y z 1 
xi yi zi 1 m 
x% y% zt 1 
x% y% z% 1 



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§232] 



SOLID ANALYTIC GEOMETRY 



293 



points on the surface lies wholly in the surface, and since 
this property is characteristic of the plane alone 

Ax + By + Cz + D = 

is the equation of a plane. 

232. Normal form of the equation of a plane. — Let the 
length of the perpendicular OR, Fig. 190, from the origin to 
the plane be p, and let its direction angles be a, 0, 7. If P is 
any point in the plane, the projection of OP on OR will be 
constant and equal to p. By theorem 2, Art. 209, the projection 
of OP equals the sum of the projections of the broken line 
ONMP on OR. 

Therefore proj. ON + proj. NM + proj. MP =* p. 
But proj. ON on OR = x cos a, 

proj. NM on OR = y cos 0, 

proj. MP on OR = z cos 7. 
Substituting these gives 
[68] x cos a + y cos ff + z cos y = p. 

This is called the normal 
form of the equation of a 
plane. 

In article 231, it was shown 
that every equation of the 
first degree in x, y and z is 
the equation of a plane. This 
article proves the converse of 
that theorem, namely, that 
every equation of a plane is 
of the first degree in x> y y 
and z. 

233. Reduction of the 
equation of a plane to the normal form. — The equations 

Ax + By + Cz + D = and 
x cos a + y cos ft + z cos 7 — p = 




Fig. 190. 



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294 ANALYTIC GEOMETRY [§234 

will be the equations of the same plane if they differ only by 
a constant factor. Suppose that k is such a factor, then 

kAx + kBy + kCz + kD - 0. (1) 

And therefore kA = cos a, 

kB = cos j8, 
kC = cos 7. 

Squaring each equation and adding gives 

k 2 (A 2 + B 2 + C 2 ) = 1. 

<* * = ± y / A 2 +B * + & 

Substituting this value of k in equation (1), gives 



Ax + By + Cz + D 

±Va* + b* + c* 



where 



COS a = 



±Va 2 + b 2 + c 2 



[59] o B 

1 J cos 3 = , ^ > 

±Va 2 + b 2 +c 2 
c 

cos y 



±VA 2 + B 2 +C 2 
-D # 

p ±Va 2 + b 2 + c 2# 

234. Intercept form of the equation of a plane.— -Let the 

plane cut the s-axis in the point where x = a, the y-axis in the 
point where y = 6, and the z-axis in the point where z = c. 
These three quantities are called the intercepts of the plane 
on the axes. If they are given and none of them is zero, the 
plane is uniquely determined, for this is equivalent to giving 
the three points on the plane (a, 0, 0), (0, 6, 0), (0, 0, c). To 
find the equation of the plane, substitute these three coordi- 
nates in succession in the general equation 

Ax + By + Cz + D = 0. (1) 



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§235] SOLID ANALYTIC GEOMETRY 295 

This gives the three equations: 



Aa + D = 0, 
Bb + D = 0, 
Cc + D = 0. 
From which 

a o 


c 


Substituting these values in (1), 




a o c 




Dividing by — D, 




°> i+i+S-«- 





(2) 



This is called the intercept form of the equation of a plane. 
Note that this form is not valid if any of the intercepts 
are 0, that is if the plane passes through the origin. 

235. Equation of a plane determined by three conditions. 
The general equation of a plane, 

Ax + By + Cz + D = 0, 

involves four constants, A, B, C, and D. Any three con- 
ditions that determine a plane give three relations between 
these four constants. These three equations can be solved 
for three of the constants in terms of the fourth providing that 
the fourth is not zero. Then after substitution the equa- 
tions can be divided through by the fourth constant as in 
equation (2), Art. 234. 

If the fourth constant should be zero, the three equations 
will turn out to be inconsistent. In such a case solve the 
equations in terms of another constant. 

236. Angle between two planes. — Let the two planes be 

Aix + Biy + dz + Di = 0, 
and A t x + B 2 y + C 2 z + D t = 0. 



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296 ANALYTIC GEOMETRY [J237 

The angle 6 between these two planes is the angle between 
their normals. Hence by [66] and [69], 

[61] cos 6 = AxA, + BA + C1C 

± VAx* + Bx* + d* V A,* + B,« + <V 
The two planes are perpendicular to each other if 

AiA t + BiBi + CiC, - 0. 
The two planes are parallel if their* normals have the same 

direction cosines, that is. if ~r = ^ = tt' 

A% D% Kj% 

237. Distance from a point to a plane. — Let Pi(xi, yi, z x ) 
be the given point and Ax + By + Cz + D = be the given 
plane. Pass a plane through Pi parallel to the given plane, 
and find the difference between the normals to the planes. 
It is then found that the distance d is given by the formula 

[62] d .A»i + 3ri + tei + D 

±Va* + b 2 + c* 

where the sign is chosen to make 4 positive. 

Example 1. — Find the equation of a plane passing through the points 
(2, 1, 7) and (4, —1,-2) at a distance 2 from the origin. 
Use the normal equation of a plane, 

x cos a+ y cos 4- * cos y — p « 0, 
Since the distance of the plane from the origin equals 2, 

p=2. 

Since the plane passes through the points (2! 1, 7) and (4, —1, —2). 

2 cos a + cos + 7 cos y — 2 »» 0, 

4 cos a — cos — 2 cos y — 2 «■ 0. 

Solving these equations with the identity 

cos* a 4- cos* 4- cos* 7 — 1, 
gives cos a = $, cos = — f, cos 7 « f, 

or cos a = H , cos = Hi cos 7 - -rh- 

Therefore, there are two solutions to this problem and they are 
Zx - 6y 4- 2z - 14 - 0, 
and 105s 4- H4y - 2z - 310 = 0. 



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J237J SOLID ANALYTIC GEOMETRY 297 

Example 2.— Find the equation of the plane bisecting the angle between 
the planes 3s — y + 2z — 4, and 2x + dy — z — 4. 

If P(x, y) is any point on the bisecting plane, its distance from each of 
the two planes is the same. Equating these distances gives 
3s - y + 2z - 4 2x + 3y - z - 4 
±Vl4 " ±Vl4 

This gives the two planes 

x - 4y + 3* -0, 
and 5s+2y +2-8-0. 

EXERCISES 

Write the equations of the planes in exercises 1-4 in the normal 
form and the intercept form. 

/ 1. x - 2y - 2« - 4. 8. 4s + 7y - 4z + 3 - 0. 

2. 2x + V - 2z - 9. 4. 12s - y + 12s - 18. 

Find the equations of the planes which satisfy the conditions of the 
exercises 5 to 16. 

5. Passing through the points (1, 1, 1), (-3, 3, 8), (-2, -3,-2). 

6. Passing through the points (2, -1, 0), (4, —2, 4), ( — 1, 3, -1). 

7. p = 5, cos a - J, cos = — }. 

8. a =» J, b = -}, c - 2. 

9. Passing through the points (4, 0, —1), (6, 3, 3) at a distance 2 
from the origin. 

10. Passing through the point (1, —2, 1) and parallel to the plane 
y- Sx +4« - 5 = 0. 

11. Passing through the points (1, 1, 1), (2, — 1, 2) and perpendicular 
to the plane 3x + 4y - 7z + 10 = 0. 

12. Passing through the point (—2, —1, 3) and perpendicular to each 
of the planes 2x — 2y — 7z + 3 =» 0, and 4x + y — 4z — 1=0. 

13. Passing through the point (1, 1, 2) and perpendicular to the line 
joining (3, -4, 2) to (4, -6, 3). 

14. Perpendicular to the line joining (7, —6, 3) to (1, 2, —5) at its 
middle point. 

15. Parallel to the x-axis and passing through the points (2, 1, 2) and 
(-3,5,5). 

16. Having the foot of the normal from the origin at the point 
(-3, 4. -2). 

17. Find the distance from the point (3, —4, 2) to the plane 
5x - 2y - 142 + 15 - 0. 



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298 ANALYTIC GEOMETRY [|238 

Find the angles between the planes in exercises 18-20. 
IB. x — y+s«7 and x + y + hz « 3. 
19. 2x + y - * « 5 and 4x — 2y — 2z « 3. 
*>. x + 2y — z - 7 and 2x — y + 7* « 10. 

Find the equations of the planes bisecting the angle between the 
planes in exercises 21-23. 

21. 2x + V — 2z«l and 3x + 6y — 2s « 7. 
22.x+V+« — 4 and 6x — y — z — 2. 

23. 2x - y - * « 3 and 5x — 5y + 2z * 4. 

24. Determine k so that fcr+0y — 7z— 22=0 shall be two units 
from the origin. 

25. Find the point of intersection of the planes 

3x + V - * - 3, 
x + 5* + 7* - 11, 
4z + lOy - 3* - -8. 

26. At what acute angle does the plane 2x + Sy + 6* = 3 cut each 
of the coordinate planes? 

27. At what acute angle does the plane 2x + 3y + 6* - 3 cut each 
coordinate axis? 

28. Prove that the planes 

2x - y + 3* = 4, 

a? + 6y - 6z - 5, 

Sx + 9y - 3« - 22, 

have a line in common. 

THE LINE IN SPACE 

238. Two plane equation of a straight line. — In article 
219, it was seen that it takes two equations in three dimen- 
sional space to define a curve. Hence the two equations 

Aix + B lV + dz + Z>x - 0, (1) 

A& + B 2 y + C* + D* = 0, (2) 

are the equations of a straight line. Equations (1) and (2) 
are the equations of any two planes through the line. 

239. Projection form of the equation of a straight line. — 
By eliminating in turn z, y, and x between equations (1) and 
(2), Art. 238, the equations of the projections of the straight 



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§240] SOLID ANALYTIC GEOMETRY 299 

line on the xy, zz, and yz-planes are obtained. If these 
equations are 

l\x + hy +h =0, 

7H\X + wi& + Ms = 0, 

niy + n 2 z + n 8 =0, 

any two of these equations are the equations of the straight 

line, and any two of these equations are called the projection 

form of the equations of a straight line. 

240. Point direction form of the equation of a straight 
line. Symmetric form. 

Case I. — The line is not parallel to any coordinate plane. 
Let Pi(xi, 2/1, Zi) be the point and let the direction of the 
straight line be given by its direction cosines, cos a, cos 
and cos 7. Then, if P(x, y, z) is any point on the straight line, 
and d is the distance from Pi to P, by [50] 

x — xi Q y — y\ z — zi , 1N 

COS a = -j — y COS fi = - - y COS 7 = -5 (1) 

Solving each equation for d and equating the results, 

[63,] *^* = *^£ = x -=*. 

1 cos a cos g cos 7 

If cos a, cos j8 and cos 7 are replaced by any quantities 
I, m, n proportional to them, the equation can be written 

[63d ^ = ^ = ^ 

1 m n 

Case II. — The line is parallel to one or two coordinate planes. 
Suppose the line is parallel to one of the coordinate planes, 
say the 2/2-plane, but is not parallel to one of the coordinate 
axes, then cos a = 0, cos 5* 0, cos 7 -^ 0. Equations [63J and 
[63 2 ] are not valid, but equation (1) can be written 

= *Jjp, cos0 = V -=^, cost = *-=p,. 

giving the equations of the line to be 

x — Xi = 0, 
y - y x z - z x 



and 



cos jS cos 7 

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300 ANALYTIC GEOMETRY [§241 

If the line is parallel to two coordinate planes, it is parallel 
to one of the coordinate axes. If this is the z-axis, then 
cos a = cos fi = 0, cos 7 = 1. Equation (1) can then be 
written 

- £^* = y -^-S cos 7 = *^ 
a a a 

giving the equations of the line to be 

x — Xi = 0, 

y - y\ = 0. 

Like considerations hold if the line is parallel to any of 
the other coordinate axes or planes. 

241. Two point form of the equation of a straight line. 

Case I. — The straight line is not parallel to any coordinate 
plane. Let the two points through which the line passes 
be Pi(xi, y x , Zi) and P%(xt f y if s 2 ). Since the direction cosines 
of this line are proportional to x% — Xi, y% — yi, and z% — Zi, 
the quantities I, m, and n of [63 2 ] can be so chosen that 

I = x t - xi, m = j/2 - yi, n = z 2 - z if 
which gives 

f64i x ~ Xi _ y-yi . «-«i 

xj-xi yt-yi zj- zr 

Case II. — The line is parallel to one or two codrdinate planes. 
The discussion is similar to that given in article 240. 

Example 1. — Reduce the equations that define the straight line, 
3s + 2y - 2z + 2 = 0, and 6x + 7y - 6s - 3 - to the symmetric 
form. 

Solution. — Reduce these equations to the projection form by first 
eliminating x and then z giving 

3y - 2z - 7 - 0, 
3* - y + 9 - 0. 

Solving each for y and equating, 

2z + 7 
3s + 9 = y - =-~ 

This can be written in the form 

x -1-3 2+1 

= y » * 



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§241] SOLID ANALYTIC GEOMETRY 301 

In order that th e denomina tors shall be direction cosines, multiply 
each equation by Vi 4- 1 + i " V* an< ^ tne equation becomes 
s + 3 j/_ * + j 

A "A" A ' 

This shows that the line passes through the point (—3, 0, — I) 
with direction cosines A, -ft, A-. If desired, the point (—3, 0, — 1) 
can be replaced by any other point on the line, say (—2, 3, 1), in which 
case the equation of the line takes the form 

a? + 2 y - 3 z - 1 

A " A " A ' 

Example 2. — Find the equation of a plane passing through the line 

2s -3 y-6 2 + 2 . 

— j — — s —= — = — g— f and the point ( — 1, — 1, —6). 

Solution. — This equation is equivalent to the two equations 
2x - 3 y-6 jtf-6 2 + 2 

—4 5- ,and -5^ 3— 

Simplifying 

lOx - 4y + 9 - 0, and Zy - 5z - 28 = 0. 
These are the equations of two planes passing through the given 
line. The equation of any plane through the line of intersection of these 
two planes, and hence through the given line, is evidently 
10s - 4y + 9 + HZy - 52 - 28) - 0. 
To make this plane pass through the point ( — 1, — 1, —6), substitute 
these coordinates and solve for k. The result is 

k = 3. 
Hence the required plane is 

10* - 4y + 9 + 3(3y - hz - 28) - 0, 
or 2x + y - Sz - 15 - 0. 

EXERCISES 

Find where the lines in exercises 1-5 intersect the three coordinate 
planes. 

1. 4x + j/+2-5=0, 2s-y + 2-l-0. 

2. x - y + z - 5, 5x - 6y + 4« = 28. 

3. 4x + j/ - 62 =10, 7* + 3y - 82 - 15 - 0. 

4. 4s + 3y + 22 - 2, -3* + 4y + 2 - 6 « 0. 
g -4 _ y + 1 _ 2-3 

•• 2 ~ -1 "" 2 ' 



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302 ANALYTIC GEOMETRY [§241 

6. Reduce the equations in exercises 1 and 2 to the projection form, 
the projecting planes being perpendicular to the xy and xz-planes. 

7. Reduce the equations in exercises 3 and 4 to the projection form, 
the projecting planes being perpendicular to the xy and j/z-planes. 

8. Reduce the equations in exercises 1-4 to the symmetric form. 
Find the equations in projection form of the straight lines in exercises 

9-18. The projecting planes are to be taken perpendicular to the xy- 
and zz-planes whenever possible. 

9. Passing through the points (3, —6, 4) and (—2, 5, 1). 

10. Passing through the points (—2, 1, 2) and (3, — 1, 4). 

11. Passing through the points (2, 1,-3) and (2, 3, —4). 

12. Passing through the points (2, 5, 6) and (2, 5, 7). 

13. Passing through the point (1, —3, 4) with direction cosines in 
the ratio 3 : — 1 : 2. 

14. Passing through the point (3, —1, 2) and parallel to the r-axis. 

15. Passing through the point (3, — 1, 2) and perpendicular to the 
z-axis. 

16. Passing through the point (3, —1, 2) and parallel to the line of 
exercise 1. 

17. Passing through the point (3, — 1, 2) and making right angles 
with the plane x — 2y + z = 3. 

18. Passing through the origin and perpendicular to the lines 

x ~ X " ~2~ ~ ^2" and ^12 " "4 3~' 

19. Find the cosine of the acute angle between the lines 

x-3 y + 2 z-6 jS-2 y-5 z 
-4- SB -=8- = -T- ,and ^ J""? 

20. Find the cosine of the angle between the line 2x — 7y — 7z = —8, 
x — 2y — z = 5, and the line 12z — 15y — 2z = 70, 5x — by — z = 24. 

21. Prove that the two lines x + y + 2 = 0, 2s — y + 3z = 7, and 
Zx + 4y + 2z = -3, -6s + 2y + lOz « meet in a point. 

22. Prove that the planes 2x + 2y + z + 4 = 0, 4s+ j/ — z — 7=0, 
and 2x + Sy + 2z + 9 = 0, meet in a straight line and find its direction 
cosines. 

Find the equations of the planes that satisfy the conditions of exercises 
23-26. 

23. Passing through the point (2, 1, 3) and the line 

3s + 5y-6z+9=0, 
2x + 2y - 2z + 1 - 0. 

24. Passing through the point ( — 1, —2, —3) and the line 

x — 1 y + 1 z + 5 
3 2 3 



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§241] SUMMARY OF FORMULAS 303 

25. Passing through the parallel lines 

x - 1 y-3 2 + 1 A x-4 y-2 2+3 
"I =7 6~ ,and -l =7- = ^6- 

26. Passing through the intersecting lines 

x y + 5 2+4 , -7x - 1 y 7«-9 
—} [ = "^- == ^9- ,and — 2 1--TT" 

27. Find the equation of a plane through the line 

x + V - 2z + 2 - 0, 
3s + 8j/ - 62 + 4 - 0, 
and perpendicular to the plane 7x + 2y + 22 — 10 = 0. 

28. Find the equation of a line lying in the plane 2x — 2y + z + 11 - 0, 
passing through the point (—3, 2,-1), and parallel to the plane 
2x + Zy - 42 + 5 - 0. 

SUMMARY OF FORMULAS 

[1] (1) OP* - OPx + PiiY (2) PxP 2 = OP 2 - OPi. 

[2] P1P2 = x% - *i. 

[2J P1P2 « s 2 - a? t , [2,] P1P2 « 1/2 - yi. 

[3] d = V(*i - x 2 ) 2 + (yi - y % )*. 

ri + r 2 ri + r 2 

[6] m - tan a = yi "" y> - 
Xi — Xa 

[7] tan <p = y-j- -• 

1 + mim 2 

[8] For parallel lines, m\ = m 2 . 

[9] For perpendicular lines, mi = > and m 2 = 

W%2 Ml 

[10] x = p cos 0, y = p sin 0, x 2 + j/ 2 = p 2 . 
[11] p = V* 2 + y\ = tan- 1 J- 

[12] x = xf + h, y =y' + k. 

[12i] a/ - * - ft, 1/ = y -k. 

[13] x = x f cos <p — y' sin $>, y = x' sin tf> + j/' cos ^>. 

[13i] x' = x cos ^> + y sin ^, y 1 = y cos ^> — x sin ^>. 



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304 ANALYTIC GEOMETRY 

[14] A = \{x x yt - 3*1/1 + x t y z - x z y t + x 9 y t - x x yt). 
[15] y - y x = m(x - Si). 
[16] y = mx + 6. 

[17] y-yi-J 1 -^ 1 («-»0. 

Xi — Xt 

[19] a; cos -f y sin tf — p =0. 
[20] Ax + By + C = 0. 

Ax By C 

roil _l y 4. = 

1 J ±VaT+b*^ ±V3h-b» t ±VI i lTB i ' 

-c 

v ~ ±va* + b* 

_ Ax x + By, + C 
1231 d_ ±VA* + 2** 

A t x + Bty + d A t x + B t y + C, 

lMi Va 1 * + b 1 * ± Vas + b,* 

[25] (x - ft)* + (V - k)* = r*. 

[26] x* + y* - r*. 

[27] x* + y* + Da; + Ey + F = 0. 

[28] y* - 2px. 

[29] x* = 2py. 

[30] (y - A)* - 2p(x - h). 

[30i] (x - *)• = 2p(y - k). 

w ' - r=V#* 



psiS+S-i- 



6* 



M fe- ffi + St- fl! . L 



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SUMMARY OF FORMULAS 
[37]y = £x, y- -£*. 



305 



o* 6* 

[39] tan2p- j4^- 
[40] * = a(0 - sin 0), y = o(l - cob 0). 
x = (a - 6) cos + 6 cob ^ a ~ ^ 9, 

y - (a - 6) sin - b sin (a T 5 ? 0. 

o 

[42] x 1 + y» = o». 

x - (a + 6) cos - b cos ^4~^ »» 



[41] 



[43] 



y - (a + 6) sin e - b sin (o ± 6) g. 



f . f x = a cos + a $ sin 0, 
^ y = a sin — a cos 0. 

[45] Axix + JBxiy + \Bxyi + Cyw + J2)x + \Dxi + 

hEy + \E yi + F = 0. 

^Jy-y^H. <*-*>• 
ax I 'as = a?i 

[47] y - yi = - ^ (3 - «i). 



da: 



s = £i 



[«] d - V(*i- x 2 ) 2 + <yt - y 2 ) 2 + (zi - *,)■. 

20 



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806 ANALYTIC GEOMETRY 

li9] *° r l + r t ' Vo ~ n + r, ' *° ~ n + r t 
[50 ]cos« = ?^- 1 , ^ fi^yJL^Jll, ^y = e -^p. 
[51] cos 2 a + cos 2 + cos 2 7 = 1. 
[52] p = ± y/x 1 + y* + z*, cos a = 



±Vx 2 + 2/ 2 + * 2 



y * 
cos = - — t=======, cos y = — 



x* + y* + z* ±Vx* + y* + z* 

[53] a: = p cos a, y = p cos 0, 2 = p cos 7. 
[54] x = p sin ?> cos B, y, = p sin <p sin 0, z = p cos <o. 



[54J p = ± V^+l/* + z\ 0= tan" 1 J' 

X 
= pit*-1 > £0 = COS -1 — — 



= sin -1 - — / . , > <p = cos" 



x* + y* ±Vx 2 + y 2 + 2 2 

[55] cos = cos ai cos a 4 + cos p\ cos /3« + cos 71 cos 74. 
[56] (x-ft)*+(y-fc) J +(2-Z)* = r 2 . 
[57] Ax + By + Cz + D = 0. 
[58] x cos a + y cos /3 + z cos 7 = p. 

[59] cos a = , > cos = , — — > 

±VA* + B* + C* ±VA* + B* + C* 

C -D 

COS "V = ; — V — , • 

±VA* + B* + C* ±VA i + B* + C* 

[60] l + l + l-l. 

[6i] cos • = ^> + **. + m 

^ + Ifo + C* + P 
±VJ r +B 2 + C* 

[63l] iLZ* = 1LZ11! = L=JL». 

L cos a cos f$ cos 7 

[63,] ^ = ^^ = i^i 

[64] *"** = V-Vi = £JZ*. 
a?j - Xi 2/2 — 2/1 «2 — *i 



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TABLES 
I. Four-place Table of Logarithms. 

II. Table of Natural and Logarithmic Sines, 
Cosines, Tangents, and Cotangents of Angles 
Differing by Ten Minutes. 



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308 



ANALYTIC GEOMETRY 
TABLE I.— COMMON LOGARITHMS 



V. 





1 


9 


3 


4 


5 


6 


7 


8 


• 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


19 


0702 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


13 


1130 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


8010 


3032 


3054 


3075 


3096 


3118 


3139 


8160 


8181 


3201 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


8444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


24 


3802 


3820 


3838 


8856 


3874 


3892 


3909 


3927 


3945 


3962 


25 


3979 


8997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 
4698 


4564 


n 


4594 
4742 


4609 


20 


4624 


4639 


4654 


4669 


4683 


4713 


4757 


80 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 
6038 


n 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 
5145 


ftP 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5159 


5172 


83 
84 


6185 
5315 


5198 
5328 


5211 
5340 


5224 
5358 


5237 
5366 


6250 
5378 


6263 5276 6289 
5391 5403 5418 


5302 
5428 


85 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


6*2t 


5539 


5551 


86 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


88 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


6877 


5888 


5899 


30 


5911 


6922 


5933 


5944 


6955 


5966 


5977 


6988 


6999 


6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


40 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


K. 





1 


2 


3 


4 


5 


6 


7 


8 


9. 



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TABLES 
TABLE I.— COMMON LOGARITHMS.— Contfmud 



309 



.-*. 





1 





8 


4 


5 





7 


8 


9 


n 


7404 


7412 


7410 
7497 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


7482 
7549 


7490 


7505 


7513 


7520 


7528 


7536 


7543 


7551 
7627 


07 


7566 


7574 


7582 


7589 


7597 


7004 


7612 


7619 


8 


88 


7642 


7640 
7723 


7657 


7664 


7672 
7745 


7679 


7686 


7694 


7701 


7716 


7731 


7738 


7752 


7760 


7767 


7774 


00 


7782 


7780 


7790 


7803 
7875 


7810 
7882 


7818 


7825 


7882 


7839 


7846 


01 


7853 


7860 


7868 


7889 


7896 


7903 


7910 


7917 


8 


7024 


7931 


7938 


7945 


7962 


7959 


7966 


7973 


7980 


7987 


7993 


8000 


8007 


8014 


8021 


« 


8035 


8041 


8048 


8055 


04 


8002 


8069 


8070 


8082 


8089 


8102 


8109 


8116 


8122 


00 


8129 


8130 


8142 


8149 


8150 


8162 
8228 


8160 


8176 


8182 


8189 


00 


8195 


8202 


8209 


8215 


8222 


8235 


8241 


8248 


8254 


07 


8201 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


00 

00 


8325 
8388 


8331 
8395 


8338 
8401 


8344 
8407 


8351 
8414 


8357 
8420 


8363 
8420 


8370 
8432 


8376 
8439 


8382 
8445 


70 
71 


8451 
8513 


8457 
8519 


8463 
8525 


8470 
8531 


8476 
8537 


8482 
8543 


8488 
8549 


8494 
8555 


8500 
8561 


8506 
8567 


7* 


8573 


8579 
8639 


8585 


8591 


8597 


8603 


8600 


8615 


8621 


8627 


73 


8833 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8680 


74 


8692 


8698 


8704 


8710 


8710 


8722 


8727 


8733 


8739 


8745 


75 


8751 


8750 


8769 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


70 


8808 


8814 


8820 


8825 


8831 


8837 


8842 
8899 


8848 


8854 


8859 


77 


8885 
8921 


8871 


8876 


8882 


8887 


8893 


8904 


8910 


8915 


78 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


00 


9031 


9080 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


01 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


89 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


88 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 
9370 


9325 


9330 


9335 


9340 


88 


9345 


9350 


9355 


9360 


9365 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


80 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


OQ 


9542 


9547 
9595 


9552 


9557 
9605 


9562 


9566 


9571 


9576 


9581 


9586 


91 


9590 


9600. 
9647 


9609 


9614 


9619 


9624 


9628 


9633 


00 


9638 


9643 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


Si 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


g 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


9823 


9827 


9832 
9877 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


97 


9868 


9872 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


S3 


9912 
9956 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


V. 





1 





3 


4 


5 





7 


8 


9 



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310 



ANALYTIC GEOMETRY 



TABLE IL— TRIGONOMETRIC FUNCTIONS 



Angles 


Sines 


Cosines 


Tangent* 


Cotangents 


Angles 




Nat. Loc 


Nat. I*c 
1 0000 0.0000 


Nat. Loc 


Nat. Log. 




(TOT 


.0000 oo 


.0000 00 


00 ' 00 


90° 00* 


10 


.0029 7.4637 


1.0000 0000 


.0029 7.4637 


343.77 2.5363 


50 


20 


.0058 76481.0000 0000 


.0058 7648 


171.89 2352 


40 


30 


.0087 9408 


1.0000 0000 


.0087 9409 


114.59 0591 


30 


40 


.0116 8.0658 


.9999 0000 


.0116 8.0658 


85.940 1.9342 


20 


50 


.0145 1627 


.9999 0000 


.0145 1627 


68.750 8373 


10 


1°<& 


.0175 8.2419 


•9998 9.9999 


.0178 8.2419 


57.290 1.7581 


89° 00' 


10 


.0204 3088 


.9998 9999 


.0204 3089 


49.104 6911 


50 


20 


.0233 3668 


.9997 9999 


.0233 3669 


42.964 6331 


40 


30 


.0262 4179 


.9997 9999 


.0262 4181 


38.188 5819 


30 


40 


.0291 4637 


.9996 WJtfo 


.0291 4638 


34.368 5362 


20 


50 


.0320 5050 


.9995 9998 


.0320 5053 


31.242 4947 


10 


2*00' 


.0349 8.5428 


.9994 9.9997 


.0349 8.5431 


28.636 1.4569 


88° (W 


10 


.0378 5776 


.9993 9997 


.0378 5779 


26.432 4221 


50 


20 


.0407 6097 


.9992 9996 
.9990 9996 


.0407 6101 


24.542 3899 


40 


30 


.0436 6397 


.0437 6401 


22.904 3599 


30 


40 


.0465 6677 


.9989 9995 


.0466 6682 


21.470 3318 


20 


50 


.0494 6940 


.9988 9995 


.0495 6945 


20.206 3055 


10 


3°00' 


.0523 8.7188 


.9986 9.9994 


.0524 8.7194 


19.081 1.2806 


87°00 f 


10 


.0552 7423 


.9985 9993 


.0553 7429 


18.075 2571 


50 


20 


.0581 7645 


.9983 9993 


.0582 7652 


17.169 2348 


40 


30 


.0610 7857 


.9981 9992 


.0612 7865 


16.350 2135 


30 


40 


.0640 8059 


.9980 9991 


.0641 8067 


15.605 1933 


20 


50 


.0669 8251 


.9978 9990 


.0670 8261 


14.924 1739 


10 


4°00 f 


.0698 8.8436 


.9976 9.9989 


.0699 8.8446 


14.301 1.1554 


86° (W 


10 


.0727 8613 


.9974 9989 


.0729 8624 


13.727 1376 


50 


20 


.0756 8783 


.9971 9988 


.0758 8795 


13.197 1205 


40 


30 


.0785 8946 


.9969 9987 


.0787 8960 


12.706 1040 


30 


40 


.0814 9104 


.9967 9986 


.0816 9118 


12.251 0882 


20 


50 


.0843 9256 


.9964 9985 


.0846 9272 


11.826 0728 


10 


5°00 f 


.0872 8.9403 


.9962 9.9983 


.0875 8.9420 


11.430 1.0580 


85° 0C 


10 


.0901 9545 


.9959 9982 


.0904 9563 


11.059 0437 


50 


20 


.0929 9682 


.9957 9981 


.0934 9701 


10.712 0299 


40 


30 


.0958 9816 


.9954 9980 


.0963 9836 


10.385 0164 


30 


40 


.0987 9945 


.9951 9979 


.0992 9966 


10.078 0034 


20 


50 


.1016 9.0070 


.9948 9977 


.1022 9.0093 


9.7882 0.9907 


10 


6° 00* 


.1045 9.0192 


.9945 9.9976 


.1051 9.0216 


9.5144 0.9784 


84° 00* 


10 


.1074 0311 


.9942 9975 


.1080 0336 


9.2553 9664 


50 


20 


.1103 0426 


.9939 9973 


.1110 0453 


9.0098 9547 


40 


30 


.1132 0539 


.9936 9972 


.1139 0567 


8.7769 9433 


30 


40 


.1161 0648 


.9932 9971 


.1169 0678 


8.5555 9322 


20 


50 


.1100 0755 


.9929 9969 


.1198 0786 


8.3450 9214 


10 


7° 00* 


.1219 9.0859 


.9925 9.9968 


.1228 9.0891 


8.1443 0.9109 


83° 00* 


10 


.1248 0961 


.9922 9966 


.1257 0995 


7.9530 9005 


50 


20 


.1276 1060 


.9918 9964 


.1287 1096 


7.7704 8904 


40 


80 


.1305 1157 


.9914 9963 


.1317 1194 


7.5958 8806 


30 


40 


.1334 1252 


.9911 9961 


.1346 1291 


7.4287 8709 


20 


50 


.1363 1345 


.9907 9959 


.1376 1385 


7.2687 8615 


10 


8° 00, 


.1392 9.1436 


.9903 9.9958 


.1405 9;1478 


7.1154 0.8522 


82*00' 


10 


.1421 1525 


.9899 9956 


.1435 1569 


6.9682 8431 


50 


20 


.1449 1612 


.9894 9954 


.1465 1658 


6.8269 8342 


40 


30 


.1478 1697 


.9890 9952 


.1495 1745 


6.6912 8255 


30 


40 


.1507 1781 


.9886 9950 


.1524 1831 


6.5606 8169 


20 


50 


.1536 1863 


.9881 9948 


.1554 1915 


6.4348 8085 


10 


9 # 00' 


.1564 9.1943 


.9877 9.9946 


.1584 9.1997 


6.3138 0.8003 


81° 00' 




Nat. Log. 


Nat. Log. 


Nat. Log. 


Nat. Log. 




Angle* 


Cosines 


Sines 


Cotangents 


Tangents 


Angles 



Digitized by 



Google 



TABLES 



311 



TABLE IL— TRIGONOMETRIC FUNCTIONS— ConHnutd 



Angles 


Sines 


Cosines , 


Tangents 


Cotangents 


Angles 


VW 


Nat. Log. 
.1564 0.1943 


Nat. Log. 
.9877 9T9946 


Nat. Log. 
.1584 971997 


Nat. Log. 
6.3138 0.8003 


81° Off 


10 


.1503 2022 


.9872 9944 


.1614 2078 


6.1970 7922 


50 


20 


.1622 2100 


.9868 9942 


.1644 2158 


6.0844 7842 


40 


90 


.1650 2176 


.9863 9940 


.1673 2236 


5.9758 7764 


30 


40 


.1679 2251 


.9858 9938 


.1703 2313 


5.8708 7687 


20 


. 50 


.1706 2324 


.9853 9936 


.1733 2389 


6.7694 7611 


10 


10* w 


.1736 9.2397 


.9848 9.9934 


.1763 9.2463 


5.6713 0.7537 


80 # 00' 


10 


.1765 2468 


.9843 9931 


.1793 2536 


5.5764 7464 


50 


£ 


.1794 2538 


.9838 9929 


.1823 2609 


5.4845 7391 


40 


.1822 2606 


.9833 9927 


.1853 2680 


5.3955 7320 


30 


40 


.1851 2674 


.9827 9924 


.1883 2750 


5.3093 7250 


20 


50 


.1880 2740 


.9822 9922 


.1914 2819 


5.2257 7181 


10 


ii° <xr 


.1908 9.2806 


.9816 9.9919 


.1944 9.2887 


5.1446 0.7113 


79° 00* 


10 


.1937 2870 


.9811 9917 


.1974 2953 


5.0658 7047 


50 


20 


.1965 2934 


.9805 9914 


.2004 3020 


4.9894 6980 


40 


90 


.1994 2997 


.9799 9912 


.2035 3085 


4.9152 6915 


30 


40 


.2022 3058 


.9793 9909 


.2065 3149 


4.8430 6851 


20 


so 


.2051 3119 


.9787 9907 


.2095 8212 


4.7729 6788 


10 


12° W 


.2079 9.3179 


.9781 9.9904 


.2126 9.3275 


4.7046 0.6725 


78*00' 


10 


.2108 3238 


.9775 9901 


.2156 8336 


4.6382 6664 


50 


20 


.2136 3296 


.9769 9899 


.2186 3397 


4.5736 6603 


40 


30 


.2164 3353 


.9763 9896 


.2217 3458 


4.5107 6542 


30 


40 


.2193 3410 


.9757 9893 


.2247 3517 


4.4494 6483 


20 


m , 


JBKM £416 


jum mm 


~£2» ISM 


4.JOT *CM 


10 


13* (W 


.2250 9 .3521 


.*744 9.9887 


.2309 9.3634 


4.3315 0.6366 


rrw 


10 


.2278 3575 


.9737 9884 


.2339 3691 


4.2747 6309 


so 


20 


.2306 3629 


.9730 9881 


.2370 3748 


4.2193 6252 


40 


30 


.2334 3682 


.9724 9878 


.2401 3804 


4.1653 6196 


30 


40 


.2363 3734 


.9717 9875 


.2432 3859 


4.1126 6141 


20 


50 


.2391 3786 


.9710 9872 


.2462 3914 


4.0611 6086 


10 


14° W 


.2419 9.3837 


.9703 9.9869 


.2493 9.3968 


4.0108 0.6Q32 


76° 00* 


10 


.2447 3887 


.9690 9866 


.2524 4021 


3.9617 5979 


50 


20 


.2476 3937 


.9689 9863 


.2555 4074 


3.9136 5926 


40 


30 


.2504 3986 


.9681 9859 


.2586 4127 


3.8667 5873 


30 


40 


.2532 4035 


.9674 9856 


.2617 4178 


3.8208 5822 


20 


50 


.2560 4083 


.9667 9853 


.2648 4230 


3.7760 5770 


10 


15° W 


.2588 9.4130 


.9659 9.9849 


.2679 9.4281 


3.7321 0.5719 


75° (XT 


10 


.2616 4177 


.9652 9846 


.2711 4331 


3.6891 5669 


50 


20 


.2644 4223 


.9644 9843 


.2742 4381 


3.6470 5619 


40 


30 


.2672 4269 


.9636 9839 


.2773 4430 


3.6059 5570 


30 


40 


.2700 4314 


.9628 9836 


.2805 4479 


3.5656 5521 


20 


50 


.2728 4359 


.9621 9832 


.2836 4527 


3.5261 5473 


10 


16° <xr 


.2756 9.4403 


.9613 9.9828 


.2867 9.4575 


3.4874 0.5425 


74° 00* 


10 


.2784 4447 


.9605 9825 


.2899 4622 


3.4495. 5378 


50 


20 


.2812 4491 


.9596 9821 


.2931 4669 


3.4124 5331 


40 


30 


.2840 4533 


.9588 9817 


.2962 4716 


3.3759 5284 


30 


40 


.2868 4576 


.9580 9814 


.2994 4762 


3.3402 5238 


20 


50 


.2896 4618 


.9572 9810 


.3026 4808 


3.3052 5192 


10 


17° W 


.2924 9.4659 


.9563 9.9806 


.3057 9.4853 


3.2709 0.5147 


73° 00* 


10 


.2952 4700 


.9555 9802 


.3089 4898 


3.2371 5102 


50 


20 


.2979 4741 


.9546 9798 


.3121 4943 


3.2041 5057 


40 


v 30 


.3007 4781 


.9537 9794 


.3153 4987 


3.1716 5013 


30 


40 


.3035 4821 


.9528 9790 


.3185 5031 


3.1397 4969 


20 


50 


.3062 4861 


.9520 9786 


.8217 5075 


3.1084 4925 


10 


18 # 0(K 


.3090 0.4900 


.9511 9.9782 


.3249 9.5118 


3.0777 0.4882 


72° W. 




Nat. Log. 


Nat. Log. 


Nat. Log. 


Nat. Log. 




Angle. 


Cosines 


Sines 


Cotangents 


Tangents 


Angles 



Digitized by 



Google 



312 



ANALYTIC GEOMETRY 



TABIA XL-TWOONOMBTRKJ FUNOITONS-CoitfftNM* 



90 
80 
40 
50 

"•ff 

90 
80 
40 
50 

20 
80 
40 
50 

*'V 

20 
80 
40 
50 

22° 00* 

10 
20 
80 
40 
50 
33° 0C 
10 
20 
80 
40 
50 

24° aor 

10 
20 y 
30 
40 
50 



20 
30 
40 
50 

20 
80 
40 
50 

WW 



3907 0.5919 
3934 5948 
3901 5978 
3987 6007 
4014 or 
4041 6005 
4067 9.0093 
4094 6121 
4120 6149 
4147 6177 
4173 6205 
4200 6T~ 

4326 9.6259 
4253 6286 
4279 6313 
4305 6340 
4331 6366 
4358 6392 

4384 9.6418 
4410 6444 
4436 6470 
4462 6495 
4488 6521 
4514 6546 

4540 9.6570 
Nat. Log. 

Coynes 



Nat. 



.9455 
,9446 
.9436 
.9426 
.9417 
.9407 

.9397' 

.9387 

.9377 

.9367 

.9356 

.9346 

.was 

.9325 
.9315 
.9304 
.9293 



.9239 



.9076 « 

.9063 9.9573 

,9051 9567 

.9038 9561 

.9026 9555 

.9013 9549 

,9001 9543 

.8988 9.9537 
.8975 9530 
.8962 9524 
.8949 9518 
.8936 9512 
.8923 9505 
.8910 9.9499 
Nat. Log. 

Sines 



TanffenU 




Cotangent* 

Nat. Log 

8.0777 0748L 

3.0475 4839 



*3' _ fc ._ 

,3541 ^491 

.3574 5531 

,3607 6571 

9.5611 




3.0178 
2.9887 
2.9600 
12.9319 

19.9042 0.4630 
2.8770 4589 
8502 4549 
_.&239 4509 
2.7980 4469 
3.7725 4429 

8,7475 0.4389 
2.7228 4350 



2.6985 
9.6746 
2.6511 
2,627$ 

2.6051 0.4158 
2.5826 41 
2.5605 



2.5386 
2.5172 
2.4960 



or 

.4314 
.4348 
.4383 
.4417 



.4040 9.6064 
.4074 6100 
.4108 6136 
6172 
6208 
6243 

9.6279 
6314 
6348 
6383 
6417 
6452 

.4452 0.6486 
.4487 6520 
.4522 6553 
.4557 6587 
.4592 6620 
.462a 6654 

.4663 9.6687 
.4699 6720 
.4734 6752 
.4770 6785 
.4806 6817 
.4841 6850 

.4877 9.6882 
.4913 6914 
.4950 6946 
.4986 6977 
.5022 7009 
.5059 7040 

.5095 9.7072 
Nat. Log. 

Cotangent! 



4046 
4009 
3972 

2.4751 0.3936 
2.4545 3900 
3864 



4797 
4755 
4713 
4671 



4311 
4273 
4234 
4196 




,.31_ 
2.2998 
2.2817 
2.2637 
2.2460 0.3514 
2.2286 3480 



3447 
3413 



2.2113 
2.1943 

2.1775 „„ 

2.1609 3346 

2.1445 0.3313 

2.1283 3880 

2.1123 3248 

2.0965 3215 

2.0809 3183 

2.0655 3150 

2.0503 0.3118 
2.0353 3086 



2.0204 
2.0057 
1.9912 
1.9768 



3054 
3023 
2991 
2960 

1.9626 0.2928 
Nat. Log. 

Tangents 



40 
30 
20 
10 

40 
30 
20 
10 

70°00 r 
50 
40 
30 
20 
10 

40*00? 
50 
40 
30 
20 
10 

68° 00* 
50 
40 
30 
20 
10 

67° 00* 
50 
40 
30 
20 
10 

66° 00? 
50 
40 
30 
20 
10 

65° 00? 
50 
40 
30 
20 
10 

64*00? 
50 
40 
30* 
20 
10 

63* c*y 



Digitized by 



Google , 



TABLES 



313 



TABUI n.—TRiaONOMETRIO TmiCnom-ConHmu* 



A*gi* 


Ones 


Cosines 


Tangents 


Cotangents 


Aaglss 


iroff 


.*W 076&0 


%*• Loft 
.8910 979499 


Nat. Log. 
.6095 9.7072 


Nat. Lag. 
1.9626 0L2928 


68*0ff 


10 


.4566 6595 


.8897 9492 


.5132 7103 


1.9486 2897 


50 


20 


.4592 6620 
.4617 6644 


.8884 9486 


.5169 7134 


1.9347 2866 


40 


80 


.8870 9479 


.5206 7165 


1.9210 2885 


80 


40 


.4648 6668 
.4669 6602 


.8857 9473 


.6243 7196 


1.9074 2804 
1.8940 2774 


20 


50 


.8848 9466 


.5280 7226 


10 


«raor 


.4698 0.6716 
.4720 6740 


.8820 0.9459 


.5817 0.7257 


1.8807 0.2743 


62* OOP 


10 


.8816 9453 


.5354 7287 


1.8676 2713 


50 


20 


.4746 6768 


.8802 9446 


.5392 7317 


1.8546 2Q83 


40 


80 


.4772 6787 
.4797 6810 


.8788 9439 


..5430 7348 


1 .8418 2652 


30 


40 


.8774 9432 
.8760 9425 


.5467 7378 


1 .8291 2622 


20 


50 


.4828 6888 


.5505 7408 


1.8165 2592 


10 


iroff 


.4848 0.6856 


.8746 0.9418 


.8548 0.7488 


1.8040 0.2562 
1.7917 2533 


61° Off 


10 


.4874 6878 


.8732 9411 


.5581 7467 


50 


20 


.4899 6901 


.8718 9404 


.5619 7497 


1.7796 2503 


40 


80 


.4924 6923 


.8704 9397 


.5658 7526 


1.7676 2474 


30 


40 


.4950 6946 


.8689 9390 
.8676 9383 


.5696 7556 


1.7556 2444 


20 


50 


.4975 6068 


.5735 7585 


1.7437 2415 


10 


*V 


.5000 0.69M 


.8660 0.0375 


.5774 0.7614 


1.7821 0.2386 


60° Off 


.5025 7012 


.8646 9368 


.6812 7644 


1.7205 2356 


50 


20 


.5050 7033 


.8631 9361 


.5851 7673 


1.7090 2327 


40 


80 


.5076 7055 


.8616 9353 


.5890 7701 


1.6977 2299 


30 


40 


.5100 7076 


.8601 9346 


.6930 7730 


1 .6864 2270 


20 


50 


.5128 7097 


.8587 9338 


.5960 7769 


1.6753 2341 


10 


81° Off 


.5180 0.7118 
.5175 7139 


.8572 0.9331 


.6009 0.7788 


1 .6643 0.2212 
t.6534 2184 


59° Off 


10 


.8557 9323 


.6048 7816 


50 


20 


.5200 7160 


.8542 9315 


.6088 7845 


1.6426 2165 


40 


80 


.5225 7181 


.8526 9308 


.6128 7873 


1.6319 2127 


30 


40 


.6250 7201 


.8511 9300 


.6168 7902 


1.6212 2098 


20 


50 


.5276 7222 


.8496 9292 


.6208 7930 


1.6107 2070 


10 


Wtt' 


.8290 0.7242 
.5324 7262 


.8480 0.9284 


.6240 0.7958 


1.6003 0.2042 


58° Off 


10 


.8465 9276 


.6289 7986 


1.5900 2014 


50 


20 


.5348 7282 
.5378 7302 


.8450 9268 


.6330 8014 


1.5798 1986 


40 


80 


.8434 9260 
.8418 9252 


.6371 8042 


1.5697 1958 


30 


40 


.5398 7322 


.6412 8070 


1.5597 1930 


20 


50 


.5422 7342 


.8403 9244 


.6453 8097 


1.5497 1903 


10 


W°0ff 


.8446 9.7361 


.8387 0.9236 


.6494 9.8125 


1.5399 0.1875 


57° Off 


10 


.5471 7380 


.8371 9228 


6536 £153 


1.5301 1847 


50 


20 


.5495 7400 


.8355 9219 


.6577 8180 


1 . 5204 1820 


40 


30 


.5519 7419 


.8339 9211 
.8323 9203 


.6619 8208 


1.5108 1792 


30 


40 


.5544 7438 


.6661 8235 


1 .5013 1765 


20 


50 


.5568 7457 


.8307 9194 


.6703 8263 


1.4910 1737 


10 


34° W 


.5592 9.7476 


.8290 0.9186 


.6745 0.8290 


1.4826 0.1710 


56° Off 


10 


.5616 7494 


.8274 9177 


.6787 8317 


1.4733 1683 


50 


20 


.5640 7513 


.8258 9169 


.6830 8344 


1.4641 1656 
1.4550 1629 


40 


80 


.5664 7531 


.8241 9160 


.6873 8371 


30 


40 


.5688 7550 


.8225 9151 


.6916 8398 


1.4460 1602 


20 


50 


.5712 7568 


.8208 9142 


.6950 8425 


1.4370 1575 


10 


WOO* 


.5736 9.7586 


.8192 0.9134 


.7002 0.8452 


1.4281 0.1548 


55° Off 


10 


.5760 7604 


.8175 9125 


.7046 8479 


1.4193 1521 


50 


20 


.5788 7622 


.8158 9116 


.7089 8506 


1.4106 1494 


40 


30 


.5807 7640 


.8141 9107 


.7133 8533 


1.4019 1467 


30 


40 


.5831 7657 


.8124 9098 


.7177 8559 


1.3934 1441 


20 


50 


.5854 7675 


.8107 9089 


.7221 8586 


1.3848 1414 


10 


86° 00* 


.5878 9.7692 


.8090 9.9080 


.7265 0.8613 


1.3764 0.1387 


54° Off 




Nat. Log. 


Nat. Log. 


Nat. Log. 


Nat. Log. 




Angles 


Cosines 


Sines 


Cotangents 


Tangents 


Angles 



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314 



ANALYTIC GEOMETRY 



TABLE II,— TRIGONOMETRIC FUNCTIONS— & 



Angles 


Bnm 


finsinw 


Tangents 1 Cotangents 


Angles 


M»OK 


Nat. Log. 
.1878 9/7602 


Nat. Log. 
.8000 07*380 


Nat. Log. Nat. Log. 
.7266 9.8613 1.3764 0.1387 


54? vr 


10 


.ajpl 7710 


.8073 0070 


.7310 86301.3680 1361 


50 


90 


.5095 7727 


.8066 0061 


.7355 86661.3507 1334 


40 


30 


.5048 7744 


.8030 0052 


.7400 86021.3514 1308 


30 


40 


.5072 7761 


.8021 0042 


.7445 8718,1.3432 1282 


90 


60 


.5006 7778 


.8004 0033 


•7400 87451.3351 1255 


10 


troo' 


.0018 0.7705 


.7086 0.0023 


•7536 0.877111.3270 0.1220 


53° 00* 


10 


.6041 7811 


.7000 0014 


.7581 8707 1.3100 1203 


50 


90 


.6066 7828 


.7051 0004 


.7627 88241.3111 1176 


40 


30 


.6068 7844 


.7034 8005 


.7673 88501.3032 1150 


30 


40 


.6111 7861 


.7016 8085 


.7790 88761.2054 1124 


90 


50 


.6134 7877 


.7806 8075 


.7766 800211.2876 1008 


10 


3T0CT 


.6157 0.7803 


.7880 0.8065 


.7818 0.80281.2700 0.1072 


52° OCT 


10 


.6180 7010 


.7862 8055 


.7860 89541.2723 1046 


50 


90 


.6209 7026 


.7844 8045 


.7007 8080,1 .2647 1020 


40 


30 


.6225 7041 


.7826 8035 


.7054 0006 1 2579 0004 


30 


40 


.6248 7057 


.7808 8025 


.8002 00321.2407 0068 


90 


50 


•6271 7073 


.7700 8015 


.8050 0058 


1.9423 0042 


10 


wrw 


.6208 0.7080 


.7771 0.8005 


.8006 0.0064 


1.2340 0.0016 


51*00' 


10 


.6316 8004 


.7753 8805 


.8146 0110 


1.2276 0800 


50 


90 


.6338 8020 


.7735 8884 


.8105 0135 


1.2203 0865 


40 


30 


.6361 8035 


.7716 8874 


.8243 0161 


1.2131 0830 


30 


40 


.6383 8050 


.7608 8864 


.8202 0187 


1.2050 0613 


90 


50 


.6406 8066 


.7670 8853 


.8342 9212 


1.1088 0788 


10 


lO'OO' 


.6428 0.8081 


.7660 0.8843 


.8301 0.0238 


1.1018 0.0762 


5O»0O' 


10 


.6450 8006 


.7642 8832 


.8441 0264 


1.1847 0736 


50 


90 


.6472 8111 


.7623 8821 


.8401 0280 


1.1778 0711 


40 


30 


.6404 8125 


.7604 8810 


.8541 0315 


1.1706 0685 


30 


40 


.6517 8140 


.7585 8800 


.8501 9341 


1.1640 0650 


20 


50 


.6530 8155 


.7566 8780 


.8642 • 0366 


1.1571 0634 


10 


41° <xr 


.6561 0.8160 


.7547 0.8778 


.8603 0.0302 


1.1504 0.0608 


49° 00* 


10 


.6583 8184 


.7528 8767 


.8744 0417 


1.1436 0583 


50 


90 


.6604 8108 


.7500 8756 


.8796 0443 1.1360 0557 


40 


30 


.6626 8213 


.7400 8745 


.8847 04681.1303 0532 


30 


40 


.6648 8227 


.7470 8733 


.8899 0494 1.1237 0606 


20 


50 


.6670 8241 


.7451 8722 


.8952 0519 


1.1171 0481 


10 


49° W 


.6601 0.8255 


.7431 0.8711 


.0004 9.9544 


1.1106 0.0456 


48° W 


10 


.6713 8260 


,7412 8600 


.9057 9570 


1.1041 0430 


50 


20 


.6734 8283 


.7302 8688 


.9110 9595 


1.0977 0405 


40 


30 


.6756 8207 


.7373 8676 


.9163 9621 


1.0913 0379 
1.0850 0354 


30 


40 


.6777 8311 


.7353 8665 


.9217 9646 


20 


50 


.6700 8324 


.7333 8653 


.9271 9671 


1.0786 0329 


10 


43° 00* 


.6820 0.8338 


.7314 0.8641 


.9325 0.9697 


1.0724 0.0303 


47°O0T 


10 


.6841 8351 


.7204 8620 


.9380 9722 


1.0661 0278 


50 


90 


.6862 8365 


.7274 8618 


.9435 9747 


1.0599 0253 


40 


30 


.6884 8378 


.7254 8606 


.9490 9772 


1.0538 0228 


30 


40 


.6005 8301 


.7234 8504 


.9545 9798 


1.0477 0202 


20 


50 


.6026 8405 


.7214 8582 


.9601 9823 


1.0416 0177 


10 


44 # 00' 


.6047 0.8418 


.7103 0.8560 


.9657 9.0848 


1.0355 0.0152 


46°O0T 


10 


.6067 8431 


.7173 8557 


.0713 0874 


1 .0295 0126 


50 


20 


.6088 8444 


.7153 8545 


.0770 9899 


1.0235 0101 


40 


30 


.7000 8457 


.7133 8532 


.9827 9924 


1.0176 0076 


30 


40 


.7030 8460 


.7112 8520 


.9884 9949 


1.0117 0051 


20 


50 


.7050 8482 


.7002 8507 


.9942 9975 


1 .0058 0025 


10 


^•OCK 


.7071 0.8405 


.7071 0.8405 


1.0000 0.0000 


1.0000 0.0000 


45° Off 




Nat. Log. 


Nat. Log. 


Nat. Log. 


Nat. Log. 




Angles 


Cosines 


'Sines 


Cotangents 


Tangents 


Angles 



Digitized by 



Google 



ANSWERS » 

Page 12. Art. 12. 

I. 1, 5, -8, -10, -3, 11, -16, -13. 

Pages 1*, 16. Art. 17. 

10. (5\/2, 5\/2), (10V2, 0), (5\/2, - 5V2). 

II. x - y = 0, x + ? - 0, 2x - y - 0. 
12. (0,0), (a,0), (}a, ±W§). 

18. (8, 0), (4, 4V3), (-4,4VU, (-8, 0), (- 4,-4\/3), (4, -4\/3). 

Pages 18, 19. Art. 19. 

1. (1) 15, (2) 18.385-, (3) 13.153-, (4) 16.279-. 

5. (1) 8.602+, 8.062+, 12.369+. (2) 11.402-, 8.062+, 8.062+. 

7. (3, -2) or (3, 14). 8. (1, 3). 9. (-1, 3), (-3, 5), or (13, -1). 

10. (5 + 4\/3, 6 + 3V3) or (5 - 4a/3, 6 - 3\/3). 

11. x 1 + y* - 6s - $y - 0. 

12. 5x - 7y - 26 - 0. 15. 7.550-. 

Page 20. Art. 20. 

1. 2f units to the right of P u 12 units to the left of Pi. 

2. Division point between two points and 3 in. from first. Division 
point beyond second point, 5 in. from first. 

Pages 23, 24. Art 22. 

1. (-2, 1). 2. (li, 3i). 3. (3, H). 4. (-22, 14). 6. (i, -J). 
7. (i, J), (-24, 28). 8. (1J, -1), (-li, -2i;, (-1, -4J,. 
9. (2f, 3*). 10. 10.050-, 11.180+, 12.806+. 13. (11, 14). 
14. (-1,0), (-4, -2). 

Pages 28, 29. Art. 28. 

1. (1) 1, (2) -1, (3) 1.732, (4) 0.1010, (5) £-=-£ (6) -3.1463. 

2. (1) 45°, (2) 135°, (3) 60°, (4) 5° 46', (5) tan~* £-=^ (6)107° 38'. 

3. -*. 5. 5J. 7. 6. 8. 3x - 2y - 2 » 0. 9, If. 

315 



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316 ANALYTIC GEOMETRY 

10. &r - y + 12 - 0. 11. x + 2y - 11 = 0, (-3, 7). 12. 60° 15'. 

18. 2.375. 14. 86° 11'. 15. lOf. 16. 3.732. 

19. (lft, 3ft). 20. 0.6584. 21. 74° 56'. 

Pages 82, 83. Art 80. 

1. (3\/2, 3V2,) (-V2,- V2), (-1, -|V3), (3^3,3), 
(-t,fV3),(-4 % /2 f 4 V ^2;,(-2, 0), (0,-6). 

2. (1.532, 1.286), (1.026, 2.819), (5.629, -3.250), (-0.7714, 0.9192), 
(4.078, -1.902). 

3. (8, 60°), (-8, 240°); (3V2, 225°), (-3<\/2, 45°); (V34, $9 Q 2')» 
(- V§4, 239° 2'); (2V2, 120°), (-2V5, 300°). f. 4.58. 

Pages 35, 36. Art 88. 

2. (f\/2, JV2),(iV2, |\/2), (-5V2, 0),(3\/2,-4\/2). 

3. (0, 0), (9, 2), (5, 11). 4. (7, 8). 

5. 60°; (0, 0), (4, 0), (2 Wl. V^8 + %). 

Pages 38, 8*. Art 85. 

1. (1) 76, (2) 31, (3) 200}, (4) 10. 2. 160. 8. 18. «. 72. 

. 7. i\pu>t sin (0i - Ox) + pips sin (0* - t ) + a»pi 8in fo — #i)]. 
8. 98.29. 

Page 41. Art. 86. 
1. (-34,2). 

Pages 41-43. General Exercises, 
2. (0, 0), (8, 0), (0, 10), (^8, 10). 3. (-8, 0), (0, 0), (8, 10), (0, 10). 
4. (4, 0), (0, 4), (-4, 0), (0, -4). 5. (a + iaVs, Ja), (}aV3,|a). 
6. (0, -*a\/3), (}a, *a\/3), (-*<*, |aV3) or (0, *aV3), 
(ia,-i«V3), (-K -iaV3). 7. 4.799. 8. 2:5. 
9. (6, -4), (14, -20). 10. (0, 9), (3, 0). 11. (1, 4). 
12. «, V). 14. (7, 0) or (-2, 0). 15. 6 or -2. 
16. (5, 0), (0, -1). 17. (5, 0), (-3, 4). 19. -}. 2a -^1.128. 
21. -0.145. 22. 1.2337. 23. 2.25. 24. (0, 0), (4ft, lit), (0, 13), 
(-4ft, lift). 25. (0, 0), (5,0), (4, 3), (-1, 3). 

Page 49. Art. 44. 
1. 5, -3*. 2. ±6, ±6. 3. ±4, ±8. 4. 1 ± iVZ, 1. i. 0, a 

6. 0, 2 and 0. 7. -2, 1, 3 and ±\/6. 8. None. 



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ANSWERS 317 

*agtl4. Art. 43. 

JL *V -* 3* - 0. 8. *» — 4*? + 4y* - 9 - 0. 

3. *» - 8** + 15* - 0. 4. *• - x*y + *y* - y« - 16* + 16* - 0. 

5. x«y - te 1 - 4*y + *y* - 6y* + 24 » 0. 

Page 86. Art. 49. 

I, (2V2, — 2V5), (— 2\/2,2x/2). 2. (t\/2, JVTi), (J\/2, -J\/l4). 
(-|V^,iVl4),(-!v^, WTi). 8. (24fc -12?), (3, 2). 

4. (fVlO, lVT5,, (fVlO,- fVifi), (- JVT6, *Vl6), 
(-jVlO, -*\/l5). 

5. (3, 4), (3, -4). 6. 4^2. 

Pages 67, 68. Art. 61. 

II. 22* + 120 - 1 - 0. 12. ** + y» + 12* + 16y «0. 

13. *» + y % - 6* - Sy - 0. 14. ** + y» + 2* + 4y - 20 * 0. 
16. 18* + 20y + 51 - 0. 

16. 4* + 3y - 25 - 0, 2* - 6? + 29 - 0, 3* - y + 2 - 0. 

17. 4* - y « 0. 18. 8** + 8y» + 112* - 30y + 347 - 0. 
19. 2* - 3y + 24 - 0. 20. 7*» + 16y» - 112 - 0. 

21. 5** - 4y» - 20 - 0. 22. 2* - 6* - 5 - 0, 2* - 6y - 15 - 0. 
23. xy + 7* + 8y - 4 - 0. 

Page 61. Art. 64. 

1. * - 2y - 8 = 0. 2. * + y + 6 - 0. 

3. \/3* + y - VS - 5 - 0. 4. 3* - by + 13 - 0. 

6. 5* + 3y - 1 - 0. 6. * - 3 - 0. 

7. y - 4 ' - 0. 8. * - 2y + 5 = 0. 

9. 3* - 4y - 11 - 0; 3* + 4y + 5 - 0. 10. 12* - by - 26 = 0. 

It. 11* - y - 16 - 0. 12. 7* - y - 5 - 0. 

Page 62. Art. 66. 
1. * - 2y + 6 - 0. 2. 3* - y - 2 = 0. 

3. 2* - 3y + 1 =* 0; 2* + 3y - 1 - 0. 4. 2* + 3y - 20 = 0. 

6. * + y + 1 - 0. 6. * - 3y - 5 =* 0. 

Page 64. Art. 68. 

1. 2* - Sy - 6 - 0. 2. 6* ' - y + 6 ~ 0. 

3. 4* + 3y - 2 - 0. 4. 3* + 4y + 1 - 0. 

6. * + V3y -6=0. 6. * + y/Zy + 6=0. 

7. * - y + 2 V2 =0. 8. \/Sx + y + 2 - 0. 
9. \/3* - y + 8 - 0. 10. V3* - y - 4 - 0. 



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318 ANALYTIC GEOMETRY 

Pages 67, 68. Art 62. 
1. -f, 2. 2. |, |. 8. |, 3. 4. 1, }. 5, *. 6. 2. 7. 2. 8. 3/ 

9.?^?. 10. !b£I?. 11. scosO° + ysinO° + i «0,i. 
5 5 

12. s cos 90° + y sin 90° - | - 0, |. 18. 3s - 4y + 5 = 0. 

14. x + 2y + 4 - 0. 15. x - 2y - 2 - 0. 16. * + y + 2 = 0. 

Page 69. Art 68. 

1. f . 2. 1. 8. 1.4142. 4. 0.232. 8. 1.828. 6. 4.427. 
7. 2, Jf, ft. 8. 5.233, 6.871, 3.757. 

Page 71. Art. 64. 

1. x - Zy - 3 = 0. 2. 2s + 6y - 7 « 0. 

3. x - 7y + 42 - 0. 4. 16s - 4y - 17 - 0. 

5. 30s + lOy + 9 - 0. 6. 2x + y - 2 - 0. 

7. 7s - 9y - 0, 4s + 6y - 21 - 0, 5s + y - 14 - 0. 

8. x + Zy - 4 = 0, x - 7y - 19 = 0, 2s - 17 - 0. 

Page 72. Art. 65. 

I. mx - y + 2m + 3 = 0. 2. y - mx = 0. 

8. ma; - y - Zm + 4 - 0. 4. | + | - 1. 

5. mx — y — 4 = 0. 6. s cos -f y sin -r- 3 = 0. 

7. s cos -f- 2/ sin - 7 « 0. 8. 2s -f- y - b = 0. 

9. 3s + y — 6 — 0. 10. s + y — 6 =* or y = »w. 

Pages 74, 75. Art. 66. 

1. 3s + 4y ± 6 = 0. 2. s - 2y - 0, s + y - 6 - 0. 

8. 3s + 4y + 5 - 0, 5s + I2y - 13 - 0. 

4. y - 2 - 0, 4s - Zy - 10 = 0. 5. 2s - y - 6 - 0. 

6. 2s + y - 6 - 0. 7. s + 4y - 4 - 0. 8. s - y - 3 = 0. 

9. s + y - 4 - 0, 3s + y - 6 - 0. 10. s + 3y - 6 = 0. 

II. s + VZy -6=0. 12. 3s - 4y ± 10 = 0. 

13. 4s - Zy + 15 - 0, 4s + Zy - 15 - 0. 

14. 3s + 4y ± 10 - 0, 4s + Zy ± 10 - 0. 

15. 2s + y - 4 - 0. 

16. 2s - (3 - 2y/2)y + 4 - 4\/2 = 0, 
2s - (3 + 2V2)y + 4 + 4\/2 - 0. 



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ANSWERS 319 

Page 77. Art. 67. 

1. 4x - by + 1 = 0. 2. 13s + \2y - 62 - 0. 8. x - 2y + 6 - 0. 
4. 22a; + lly - 14 - 0. 5. 5a? + I5y - 34 - 0. 

Page 78. Art. 68. 

1. x*y — y* - xy + y* = 0. 

2. *V ~ 3s*y - 3xy* + 2x» + 9xy + 2y» - 6a: - 6y + 4 - 0. 

Page 79. Art. 69. 

1. p cos (0 - 45°) =3. 2. p cos (0 - 60°) - -2. 

3. p sin — 7. 4. p cos 6 = —4. 

5. p cos (0 - 135°) = -4. 6. p cos (0 - 315°) - 3. 

7. p cos (0 - 45°) = J V2. 8. p cos = 3. 

9. p sin - -7. 10. p cos (0 - 30°) = 2. 

11. p cos (0 - 300°) = -3. 12. tan 0=2. 
13. x - 3 - 0. 14. y - 4 - 0. 

15. y - 6a: - 0. 16. x + y - 2 = 0. 

17. a: + y - 3 = 0. 18. 4a; - 6y - 3 - 0. 

19. 3a: ± 4y = 0. 20. 12a: ± 5y = 0. 

21. a: - y + 2 = 0. 22. x - VSy -6=0. 

Pages 81-85. General Exercises. 

2. (1) 2a: - Zy - 17 = 0. 3. (1) x - 7y - 33 = 0. 
(2) 7x - 2y - 21 = 0. (2) y = mx + k - mft. 

4. (1) i. 

(2) 0.5883. 

(3) Vj_ 

5. f i-v/82, fi VSI, V V5. _ 

6. AVIS, H Vl3, if Vl3, A Vl3, A Vis. 

7. (1) 7. 8. (1) a: + Sy - 58 = 0. 
(2) -f. (2) 21a: - 6y - 58 = 0. 

(3) 12a: + 9y - 116 = 0. 

(4) 15a: + 6y - 110 = 0. 

9. 3a: - y - 1 = 0, 7a; + by - 15 = 0, x - 4y + 6 = 0, (H, H). 

12. a; + V3y -4=0. 13. (1) x - y + 2 = 0. 

(2) 2a: - y - 1 = 0. 

(3) V3a: + y - 8 + 2 V3 = 0. 
15. 2a; - jy + 6 = 0. 16. («, J). 17. a;' - by'+ 4\/2 = 0. 
19. 4a; - by + 1 - 0, 7a: - Sy - 27 = 0, 3a: + 2y - 5 - 0. 



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320 ANALYTIC GEOMETRY 

20. s - y - 1 - 0. 21. s + y - 8\/2« 0. 

22. 3* + 4y - 20 - 0. 23. 7x - 4y - 2 - 0. 

24. x - 3y + 7 - 0, 13* + 9y - 5 - 0. 25. x - 2y + 2 - 0. 

26. x - (2 - \/3)y -1 + vl -0,«-(2 + V3)v - 1 -VI - 0. 

27. 3s - 4y + 24 - 0. 28. 3s + 4y - 24 - 0. 
29. x - y - 5 - 0, a? + y - 13 - 0. 80. -2.0225. 
31. 3. 32. x - 2y + 7 - 0, x + 3y - 8 » 0. 

83. s-(2+\/3)y+ll+6\/3«0, s-(2-\/3)y-4- v"3 - 0. 

34. x + 3y - 15 - 0, Zx - y - 5 - 0. 

88. *\/5, (1, V). 36. (3, 4). 37. 2x - y + 4 - 0. 

88. 3s - 4y - 0. 89. 3s + Zy - 13 - 0, 3s + Zy - 11 - 0. 

40. 3s - 4y + 1 - 0, 3s + 4y - 7 - 0. 

42. 4s - Zy + 6 - 0, 3s - 4y + 15 - 0. 

44. s- V3y - 0, s + \/3y = 0. 45. s - y + 1 - 0. 

46. s - VZy - 0. 47. y = 4, y - 3, s = 2, s « 3. 

48. (^, ^} . 49. s + (8 - 5 \/3)y - 50 + ZOVz - 0. 

50. (a -f b)x — (a — 6)y — be — ac - 0. 

51. 3s 7 + 5y' + 4 = 0. 

52. 21s + 77y - 1 - 0, 99s - 27y - 79 - 0. 

54. (1) A - £ or C =* 0. 55. (1) 116° 84\ 

(2) A » -£. (2) 79° 42'. 

(3) A+ 2B + C - 0. (3) 60 d . 
56. 8 + 6\/3. 57. (-1, 1). 58. 3. 

59. 8s - by + 30 - 0. 60. 175° 26'. 

61. 18s + 129y - 50 - 0, 138s + 79y - 210 - 0. 

62. s + Zy - 30 = 0, s + Zy + 10 - 0. 63. y + 6 - 0. 

64. (2, 4). 65. s + y - 2 = 0, s - (2 + V3)y - 2 - 2\/5 - 0, 

s - (2 - V3)y - 2 + 2V§ * 0. 









Page 88. 


Art 74. 








1. 

5. 
7. 
9. 


(l, 2); 
(i 1); 
<-*, ■ 

(-3a, 


3. t. (- 

iV2- 

- 1); o. 

-ia);iV7, 


2, 3); 1. 
i. 


8. (-6, -J 

e. (-i, 

8. (a, 3a) 
10. (Ja, - 


J); 2. 4. 
-i);iV2. 
; 3a. 

|a);Ja. 


0, 2); 


i 



Pages 90-92. Art. 75. 

1. s* + y* - 6s - &y + 20 = 0. 2. s* + y* - 2s + 6y + 5 - 0. 
3. s* + j/» + 4s - 4y + 3 - 0. 4. s» + y* - 2s - 2y - 23 - 0, 



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ANSWERS 821 

5. a;* + y* + 2x + 2y - 23 - 0. 6. a* + y* - 4a: - 6y - 12 - 0. 

7. a;* + y* - 2y - 12 - 0. 8. a;* + y* - 2x - 24 « 0. 

9. ** + y* - 4a? - 8y + 10 - 0. 10. a;* + y* + 4a; - Sy + 10 - 0. 
11. a* + y* ± 8a; + 6y - 0. 12. a;* + y* - 24a; ± lOy - 0. 

13. a* + y* - 20. 14. ** + y* - 2» - 4y - 8 - 0. 

15. x* + y* + 2a; - 6y = 0. 16. a;* + y* - 4a; - 6 - 0. 

17. x* + y* - 4a; - 4y - 17 - 0, a;* + y* - 10a; - 22y + 121 - 0. 

18. a;* + y* + 6a; - 2y - 15 - 0, a;* + y* - 10a; - 14y + 49-0. 

19. x* + y* - 2a; - Sy - 3 » 0. 20. a;* + y* + 4a; - 6y - 0. 

21. a;* -f y* - 2a; - 2y - 3 = 0. 22. a:* + y* - 6a; ± Sy + 9 - 0. 

28. a;* + y* ± 4a; - 8y + 16 - 0. 24. x* + y* - 4a; - Sy + 10 - 0. 
25. a;* -f y* - 6a; - 6y + 9 - 0. 26. a?* + y* - 2a; + 2y + 1 - 0. 
27. a;* + y* + 4a; - Ay + 4 - 0. 28. a;* + y* + 6a; - 6y + 9 - 0. 

29. a;* + y* - 4a; - 4y - 2 - 0. 80. ** + y* + 4c - 6y + 8 - 0. 
31. 4a;* + 4y* + 20a; - 20y + 25 - 0, 

x 1 + y* + 30a; - 30y + 225 - 0. 

82. x * + y* - 10a; - lOy + 25-0. 

83. a;* + y* - 10a; - lOy + 25 - 0, 
x* + y» - 26a; - 26y + 169 - 0. 

34. 2a; - y - 7 - 0. 85. * - 2y + 5 - 0; 2V& 

86. x - 3y + 5 - 0; 2\/l0. ' 87. x* + y* - 85 - 0. 

88. 36a;* + 36y* + 84a; - 12y - 575 - 0. 

89. a;* + y* - 65. 

Page 94. Art. 76. 

1. x + 2y - 5 - 0. 2. x + y - 1 = 0. 3. x + ?y + 4 - 0. 
4. 5a; + 3y - 7 - 0. 5. 5a; + 15y - 34 = 0. 6. 7x +9y -18 » 0. 
7. x* + y* - 3x - 4y + 5 = 0. 8. x* + y* - 5x - 13y + 42-0. 

Pages 95, 96. Art. 77. 

1. Circle, center at origin, r — 3. 

2. Circle, center (J, |), r - JvTo. 

3. Circle, center (V, 0), r = f. 

4. Circle, center (^, - 1), r = ^VOS. 

5. Two circles, centers ( ± }, 2), r - }. 6. •¥• 

7. Circle, center (0, 1), r = V29. 

8. Two circles, centers (0, ±2\/3), r = 4. 

9. Circle, center (-9, 0), r = 10. 

10. Circle, center at origin, r — 3. 

21 



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322 ANALYTIC GEOMETRY 

Page 97. Art. 78. 

1 p — 2. 2. p ~ 10 cos 0. 3. p » —8 cos 0. 4. p = 6 sin 0. 
5. p - — 4 sin 0. 6. p - ± 12 sin 6. 7. p — ± 12 cos 0. 

8. p - 6 cos (0 — i) • 9. p - VS. 10. p - 3 sin 0. 

11. p « — f cos 0. 

. 12. p = 6 cos + 8 sin j0. 

13. s* + y* + 6y = 0; (0, -3); 3. 

14. s» + y» -4s = 0; (2, 0); 2. 

15. s* + y* - s - y - 0; (J, J); JV2. 

16. s* + y* = 25; (0,0); 5. 

17. s* + y* 4- 2x + 3y - 0; (-1, - 1); iVl3. 

18. s*,+ y* + 3s + 4y - 6 - 0; (- f, -2); {. 

19. s* + y*-9 =0; (0,0); 3. 

20. s* + y*-4 = 0;(0, 0); 2. 

Page 103. Art 85. 

2. (J,0),(-i,0),(0,i),(0, -*);2s-fl.= 0,2s -1=0, 2y + 1 - 0, 
2y - 1 - 0; 2. 5. (1) y* = 12s; (2) s* - 24y; (3) y* - -16s; (4) 
s* - -8y. 6. (1) y* = 8s; (2) s* - y. 7. s* - 5000y. 

Pages 106, 106. Art 86. 

1. (1) y* - 8s - 8y + 40 - 0; (2) y* + 8s - 6y - 7 - 0; 

(3) s* + 12s - 12y + 60 •- 0; (4) s* - 4s + 6y + 22 - 0. 

2. (1) (5, 4) t s - 1 - 0; (2) (0, 3), s - 4 - 0; (3) (-6, 5), y + 1 - O; 

(4) (2, -4J), 2y + 3 = 0. 

3. y* - 10s + 4y - 36 - 0, y* + 10s + 4y + 44 = 0. 

4. s* - 6s + 8y + 25 - 0. 6. s'* + 4s' + Sy' + 32 - 0. 

6. (1) y* - Sx - lOy + 57 - 0; (2) s* + 8s - 8y + 32 - 0; 
(3) y* + 8s - 4y + 36 = 0; (4) s* - 6s + 8y + 41 - 0. 

7. (1) y* - 3s - 6y + 15 - 0; (2) 4y* + 25s - 16y - 59 - 0; 
(3) 2s* - 8s + 9y - 19 - 0; (4) 5s* - 30s - 16y + 13 = 0. 

Pages 108, 109. Art. 89. 

1. (1) (3, 2), (4, 2), y - 2 = 0, s - 2 = 0; (2) (5, -4), (4}, -4), 
y + 4=0, 2s -11'- 0; (3) (-1}, 2), (-1*, 3i), 2s + 3 = 0, 
4y - 3 = 0; (4) (6, -2), (6, -2|), s - 6 - 0, 8y + 13 - 0; 

(5) (-A, 2), (-If*, 2), y - 2 - 0, 6s - 43 - 0; (6) (|, «), (|, f|), 
2s - 9 = 0, 40y - 239 - 0. 

2. 5s* - 9s - 2y + 4 = 0, (*,-*), (,A -A),, f . 



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ANSWERS 323 

3. y* - 12* - 8y + 28 - 0. 4. 3y» - 25s + 18y + 77 = 0. 

5. (1) y" - 4s'; (2) s" - -16y'; (3) y" - -8s'; (4) s'« - Jy'. 

6. (1) 4s - 3 - 0, s' + 1 - 0, (|,3), (1, 0), 4; 

(2) y - 5 = 0, y' - 4 = 0, (4, -3), (0, -4), 16; 

(3) s - 3 - 0, s' - 2 - 0, (-1, 2), (-2, 0), 8; 

(4) 42y - 11 = 0, 12y' + 7=0, (-*, V), (0, A), J. 

Pages 110, 111. Art. 90. 

1. X 7 * - -2py', y'» = 2ps'. 

2. (1) a" + 2s'y' + y" - 4\/2s' + 4VV - 0; 

(2) 3s'» - 2\/3s'y' +y" + (6 \/3 -4)s' - (4\/3 + 6)y' + 24 -0; 

(3) 5y'« - 2\/5s' - 2\/5y' - 10 - 0; 

(4) 13s'* - 6a/13s' - 14Vl3y' - 68 - 0. 

3. 9s* - 24sy + 16y» - 116s - 162y + 321 = 0. 

4. (1) y"» - 3\/2s"; (2) s"* = AV^y"; (3) s"» - - iV2y". 



3 -*^ 



Pages 111, 112. Art 91. 



1 ± cos 

Page 118. Art. 92. 
2. (3.31, ±7.28). 

Page 114. Art. 93. 

1. s* = - 144 y, 24.31 ft., 22.22 ft., 13.89 ft. 

2. 14' J", 11' 3", 6' 6}". 3. s» - a*f*y. 

4. s» - H 8Jl (y - 20). 

Page 116. Art. 94. 
A 3t>* sin* a 

5. (1) 23.67 mi.; (2) 20.50 mi.; (3) 20.50 mi. 
7. 401.5 ft. per sec. 

Page 116, 116. General Exercises. 

1. (621, 62500). 8. 1J in. from back of reflector. 

4. (4, 2), (8, 8). 6. s - y - 2 - 0. 6. - 6}. 7. (3, 1), (42, 14). 

8. 3jt_\/57. 9 . (ff _2i). 10. (1) y» + 4s - 4 = 0; 
4 
(2) 25y 2 - 60s - 36 - 0; (3) y* + 24s - 144 = 0. 
11. (1) (12, 70° 32'), (12, 289° 28'); (2) (4, 90°), (4, 270°). 
13. 2s* + 2y» - 5ps = 0. 14. 4\/3p. 



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324 



ANALYTIC GEOMETRY ' 



Pages 122, 123. Art 9t. 

1. (1) 5,4,1, (±3,0), 3s ±25-0; 

(2) 10,6,1, (0,±8),2y ±25-0; 

(3) 3, 2, iV5, (±V5, 0), 5s ± 9V5 - 0; 

(4) 4, 3, i\/7, ( ± V7 t 0), 7s ± 16^/7 - 0; 

(5) 2V% y/l, Wl, (± V3, 0), 3s ± SVS - 0. 

(6) 3, V6, iVZ t (±V3, 0), x ± 3V3 - 0. 

2. a. 3. (1) 4s* + 9y* - 144; (2) s* + 4y* - 16; 

(3) 5** + 9y* - 81; (4) 20s* + 36y* - 1125; (5) 112** + 256y* - 3067; 
(6) s* + 4y* - 64; (7) 2s* + 3y* - 18. 

4. ± $ V21, ± 2f , 0, ± 1 V^ll. 6. 2s* + 3y* - 6. 

8. (1; \/6, VB, i V6,|\/6;(2) V2^ VJf i y/% V2^~; (3) V«, 

Jv^i, - q Vrif=D, - q Vi; (4) VJ, Vp, J Vi?Frt. f V* 

9. 3.8, 6.2. 10. 3s* + 4y* - 576. 12. (0, 0), s - ± «, 0. 

13. (1, ± 2). 14. 6*s* + c*y* - a*6*. 

15. (a* - 6»)s* + aV - a*(a> - 6*). 

Page 124. Art. 100. 

1. (1) 9s* + 25y* - 54s - 200y + 256 - 0. 
(2) 48s* + y* + 288s + 14y + 445 = 0. 

2. 4s* + 9y* - 40s + 72y + 100 = 0, 
9s* + 4y* - 90s + 32y + 145 - 0. 

3. (1) (-1, 4), (7, 4), 4s + 13 - 0, 4s - 37 - 0;' 
(2) (-3, -7 ± J VT41), y - - 7 ± H VUT. 

4. 7s* + 16y* - 140s - 64y + 512 - 0. 

5. 16s* + 25y* - 96s - 200y + 144-0. 

6. 2s* + y* - 16s - 4y - 0. 



1. (1) 



(s + 1)» 



Page 127. Art. 101. 

-2)* 



-l; (-1, 2); (^4,2), (2, 2); (-5, 2), 



16 r 7 
(3, 2); a = 4, 6 = \/7; 3s + 19 - 0, 3s - 13 - 0. 

(2) to^W+fX^Lh'-i. (4, 1); (4, 1 ±2^2); (4, -3), 
(4, 5); a - 4, 6 = 2\/2; y - 1 ± 4\/2 = 0. 

(3) ^=-^ f + &+J) 1 - l; (1 - 1); (1 ± i V5, -1); (J, -1), 

(!, -i>; o = i, 6 = J; s - 1 ± a V5 = 0. 



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ANSWERS 325 

(4) £+»'+ (JLZ.3)' , 1; (-!, 3); (-i ± J VB, 3); 

(-1 ±tV5, 3); a « $\/5, & - VI6; * + 1 ± *Vs - 0. 

2. 6s'* + 7y'* = 8. 

3. (1) *J + *P - l f (±3, 0), 3s> ± 16 « 0. 

(2) § + ^ - 1, (0, ±2 \/2), y' ± 4 V2 = 0. 



(3) 



j + j - 1, (± 1 \/5, 0), 10s* ± 3 V5 - 0. ' 



W ^ , + £ " *' (± * ^ 0) ' 2s' ± 9 \/5 « 0. 

4. 9s* + 25y* + $4s - 200y - 873 - 0; 

5. 3s* + 4y* - 24s - 16y + 16-0. 

6. 3^* ■+ 4y'* = 36. 7. 3s* + 4y* - 20s + 12-0. 

Pages 128, 129. Art. 102. 

1. (1) (a* + 6*)s'* + 2(a* - 6*)sV + («*+ W* - 2a*6* - 0, 

(2) 43s'* - 14\/3sy + 57y'* - 576 - 0; 

(3) s'* + 9y'« - 36 - 0; 

(4) 9s'« + 2 V3sy + 1 ly'* + (6 - 8 V3)s' + (6 Vs + 8)y' - 40 - 0; 

(5) 3s'« + y'* + 3-v/2s' - VV +4=0; 

(6) 38s'* + 12sV + 22y'« + 2\/5s' - 21 VV = 0. 

2. ^ + | " 1. 3- 9*"* + 3y"» -82-0. 

Page 129. Art. 108. 

1 a . e P n . gP- 4 o* ^ fl * (l - e> > 

* 1 - e ops ^ ' 1 + e cos * * 1 - e 2 cos* 

Pages 181, 182. Art. 106. 

1- ($)*+& = 1, (±m.8,0). 2. ^ + ^ = 1,(190.75,0). 

8. 36.37 ft., 27.78 ft. 8. 0.95+ ft., 4.02- ft., 10.16- ft. 

6. 13,000 mi. 

8. 45.1 in. per see. 0.14 in. per see. 

. Pages 132, 183. General Exercises. 

** 36 + 16 lm 2 * 72 + 144 " lm 5 * 72 + 36 " 1b 



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326 ANALYTIC GEOMETRY 

I 6, 2\/6; (±2a/3, 0); i\/3; x ± 6 a/3 =0. 
?. 5, 4; (-1, -2), (5, -2); {; 3x + 19 = 0, 3x - 31 - 0. 
•. 9*'» + 4y" - 36. 9. 25x" + 16y" - 400. 

x* , y* 
9 

11. 189** + 96xy + 161y» - 1494x - 25Sy + 2106 - 0. 

12. 6.4. 



10. i + ~r— n - 1. a > 3 



Pages 138, 139. Art. 110. 

1. (1) 10, 8, (±V41, 0), 41x ± 25V41 - 0; (2) 12, 20, (±2^34, 0), 
17x ± 9\/34 = 0; (3) 6, 8, (±5, 0), 5x ±19 - 0; (4) 16, 12, (0, ±10), 
5y ± 32j== 0; (5) 2V2, 2\/3, (0, ±\Z5), 5y ± 2\/5 - 0; (6) 10, 6, 
(0, ±\/84), My ± 25V34 - 0. 

3. 0, ± 5J, ± i V^5. *- 5 i~ 

5. 5x* - 9y« - 36. 6. 5x« - 4y» - 80. 

8. (1) 4, 3, J, f; (2) V6, 2y/% iV21, f V6; (3) 1, 4, Vl7, 32; 

(4) V2m, Vmj JV6, V2mj (5) Vq f Vp, ^ Vpq + q*, ~ VgJ 

9. (1) 3, 4; (± 5, 0); #; 5x ± 9 - 0. 

(2) 6, 2V6; (± 2\/l5, 0); (i Vl5; 5x ± 6\/l5 - 0. 

10. x* - 3y 8 + 3 - 0. 13. xV = 8. 14. 7x» - 9y» = 1008. 
15. x» - 3y» - 144. 

Pages 142, 143. Art. 113. 

1. (1) 2x ±V6y - 0; (2) x ±V2y = 0; (3) x ±V2y = 0; 
(4) 5x ± 4y - 0; (5) x ± y - 0; (6) x ± y - 0. 

2. 3x« - 4y« + 48 - 0, (± 2\/7, 0), (0, ±2\/7),' 7x ± 8\/7 - 0, 
7y±6V7=0. 

5. 2x'y' - a*. 6. x* - 3j/« - 16. 

11. ±4\/2. 12. ±0.9014. 



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ANSWERS 327 

Pages 148, 144. Art 114. 

1. (1) 9x* - 2by* - 72x - 150y - 306 - 0; 
(2) x« - 4y« + 12* - 16y + 36 - 0. 

2. (1) (-1, -3), (9, -3); (4 ± V34, -3); 

34x - 136 ± 25V34 - 0. 
(2) (-6, -4), (-6, 0); (-6, -2 ± 2Vb); 
by + 10 ± 2V5 - 0. 

3. 64s* - 36y* + 256x + b04y - 1283 - 0. 

4. (1) 9x» - 2by* - 72x - 150y + 144 - 0; 
I (2) x* - 4y» + 12x - lQy + 4 - 0. 

5. (1) 3x - by - 27 - 0, 3x + by + 3 - 0; 
(2) x - 2y + 2 = 0, x + 2y + 10 - 0. 

Page 146. Art. 115. 

1. (1) ^jff^* " -^T^' - 1; ». 3); (1, 3), (11, 3); (2, 3), (10,3); 
a - 4, 6 = 3; 5x - 30 ± 16 - 0; 3x - 4y - 6 = 0, 3x + 4y - 30 = 0. 
(2 ) fiLzJ) 1 . &L+i> f . i ; (-3, f); (-3, | ± Vl7); (-8,1), 

(-3, J); a - 1, 6 = 4; 34y - 85 ± 2\/i7 - 0; x - 4y + 13 ^ 0, 
x + 4y - 7 = 0. (3) &L=« f . &L+l) f = i ; (j - } ); ( _ f| _ j), 

(i, -i); (i ± \/7, -J); a - V7, & = V2; 6x - 17 - 0, 6x + 11 = 0; 
14y + 7= ±2Vl4xTVl4. (4)^^) f « &L=Li> f . 1; (1| 3); 

(1, 3 ± \/l7); (1, 3 ± 2V2); a - 2\/2, 6 = 3; 17y - 51 ± 8\/l7 - 0; 
4x - S\/2y - 4 + 9\/2 = 0, 4x + 3\/2y - 4 - 9 V2 - O.n 

(5) ^S^'"^^^ 1 ^" 2 ' 4 ^"^^ 

a - 2\/7, b - V21; y - 0, y - 8 - 0; 2x - V3y + 4 + 4\/3 - 0, 

2x + V3y + 4 - 4\/3 - 0. 

2. 9x" - 25i/'« 4- 225 =0. 

8. (1) ^- ^* - 1, (±5, 0), 5x' ±16 = 0; 

(2) *£- ~ r 1, (0, ± Vl7), 17y' ±Vl7 = 0; 

(3) y - ^ - 1, (±3, 0), 3s'±7 - 0; 



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328 ANALYTIC GEOMETRY 

(4) Y""? - *> (°» ±Vl7), MY ±SVT7 - 0. 

4. «x» - 16y» + 54* + 128y + 1601 - 0. 

5. 16x* - 25y» + 64* + 200y - 736 = 0. 

6. 3s» - y* - 84x + Ay + 536 - 0. 

Pages 14ft, 147. Art 11C 

1. (1) *Y - 8; 



(2) £*- 


b* *' 






(2) o» 






(3) x" ■ 


- y" - 16; 






(4) n*'» + sovlteV - 


■39y" - 


676; 


»?• 


3 *' 






(6) x>* 


- 3y" - 2a/ + 6/ - 11 


-0. 


(Dv- 




i; 




(2) *>« 


30« *?-£'- 
™' 3 4 


1. 





3. *"« - y" 1 - ll\/2. 

4. (-3, -1); (2.15, 1.13), (-8.15, -3.13); a? - (1 + y/2)y + 2 - 
V2 - 0, x + (\/2 - l)y + 2 + \/2 - 0; x - (\/2 - l)y + 4 - 
y/2 « 0, x + (V2 + l)y + 4 + V2 - 0. 

Paget 147, 148. Art 117. 
*• * 1 + e cos p 1 T e cos * 

Paget 149, 150. Art 118. 

4. 5x* - 4y» - 20. 

Pages 151, 152. Art 119. 

1. pv — 10. 2. t#Z = 20. 8. One branch of an hyperbola with foci 
at centers of circles and transverse axis equal to the difference of radii. 
4. Hyperbola with foci at centers of circles and transverse axis equal 
to the sum of radii. 5. An equilateral hyperbola with the ends of the 
base as vertices. 6. 345 ft. at an angle of 11° 43' with the perpendicular 
to AB. 



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ANSWERS 329 

Paget lfS, lit. General Exercises. 

1. a - \/6, b - 2; WlS; (±\/IO, 0); 3y* - 2x* - 12. 

2. (±2\/3,2),(±2\/3, -2). _ 

*. a - 6, b - 4; (3, 2 ± 2\/l3); J\/l3; 13y - 26 ± 18\/i3 - 0. 

7. (y - 2)» - 0. 

8. («V39, fjVS), (-»*V39, -«V39). 

9. 7.806+, 17.806+. 

10. s» - 3y* - x + Zy - 0. 

11. 4y» - x» - 11. 

Page 198. Art. 122. 

1. Hyperbola, 3s"» - 2y"» - 6. 2. Ellipse, 9s"* + 16^ - 144. 

3. Ellipse, 3s"» + y"» + 6-0. 4. Parabola, 2y"» - 3a;' 4 . 
5. EUipse, s" + 4y>« - 16. 6. Parabola, y'* - 3s'. 

7. Imaginary ellipse, 121*'" + lly"» + 199-0. 

8. Two lines, 2&r"» - y"* - 0. 9. Parabola, *"* - fVoy". 
10. Parabola, y"» - H Vl3s". 

Page 160. Ait. 124. 
9. Jfe - 0.0000082, p - 0.G000082P- 4 ". 10. c « 6028, *»*•* - 6028. 

Page 162, 163. Art. 126. 

17. (1) x» + y» - 2aj4nj£s + « f - °J 
( 2 )*L + y% =1- 





(3) 

p' ■ 

TT¥> 


a; 1 
Jfci" 

*2a» 

y - 


y 1 


20. 


J(4a* - Jb«) 
cos 20. 

64 2304* 



Pages 173, 174. Art 136. 

ID. 2r. 11. No. 12. y = 18 sin fcrf. 13. y - 8sin(M20° + 65°), 3, J. 
14. (1) 3.34, 0.299; (2) 16, A. 16. fcr f fcror 1 sec., 2 sec. 17. J*, 0. 
20. 0.0007854. 

Pages 179, 180. Art. 140. 

27. p - r cos $. 28. p = 2(6 + r cos 6). 



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330 ANALYTIC GEOMETRY 

Page 182. Art 14L 
l.x«-3 + *,y-2 + 2/. 2. x - 2 - 5 c<* #, y - 3 - 5 sin 0. 

4. y* + x + 2y-3 -0. §.^ + ^-1. 

6. 4x» + y» - lfix + 12 » 0. 7. x* + 2xy + y* - 2x + 2y = 0. 

8. te* + 4y«-90x-32y + 253-0. 

*. y« - tx. 10. x* - J(y + 1). 

1L x» + y* - a» + &». 12. (?) * + &\ * - 1. 

Page 183. Art. 142. 
3. x — a$ — 6 sin 0, y = a — 6 cos 0. 

Paget 193-195. Art. 161. 

1. I = 0.00102T. 2. y - 1.405x + 7.527. 

3. y - 1.403x + 7.54. 4. R = 0.00313/ + 9.8753. 

5. H - 03119/ + 606.00. 6. # = 0.1470TF + 1.7957. 
8. TF - 0.5015/ + 54.10. 

Paget 201-203. Art. 166. 

2. y - 0.9975X" 1 " or, very nearly, y = x -1 - 17 . 

3. ff - 3.867Z>>-™ 659. 4. p = 30e-o.ooooi* # 

6. m - 0.1374p-° M7 . 8. m - 0.00014V°-". 

9. pv l " - 147. 

Pages 204, 205. Art. 167. 

2. y - 0.5 + 0.02x + 2.5x* - 0.3x». 

3. / = 132 + 0.875x + 0.01125x». 

Page 210. Art. 160. 

1. x + 2y - 1 = 0. 2. 2x - 6y - 3 = 0. 

3. 3x + 4y + 1 - 0. 4. 2x + 12y + 5 - 0. 

5. 3x - 2y + 6 - 0. 6. 3x + 2y - 4 - 0. 

7. 3x + 4 - 0. 8. x + 14y + 17-0. 

9. (1, 3). 10. (5, 2). 11. (1, 2). 12. (4, 2). 13. (3, 1). 
14. (3, 6). 15. (1, 1). 16. (-1, 3). 17. (-2, -8). 

Pages 214, 215. Art. 166. 

1. x + 4y - 0, 3x - 2y - 0. 2. x + y - 0, x - 6y - 0. 

3. 3y - 2 - 0, y + 4 - 0. 4. 2x + 9y - 20 - 0. 

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ANSWERS 331 

5. 2x-9y + 16 -0. 6. (|, -*),(-!, |). 

1 o.(^^),(-^-^).i a .,-« y + 2 -o. 

13. 3x - 2y - 12 « 0. 14. 2x - 3y + 16 - 0. 

18. Ellipse concentric with original ellipse, major axis « V^2o, minor 
axis = V2&. 

P««es 218-220. Art 187. 

2. 4.5, 0.0802. 3. A = x», 13. 4. A - tx\ 10.25r. 

6. A = Jd* d - V2A. 6. C - 2\/iSf, S - ^36>F*. 

7. 7 - xr*A, h - -^ :, S - ^f. 8. -4, 44, -12. 

9. 1, VlO, 7. 10. 2, 1.6778, -1.9208. 11. iV3, 0, -J. 
12. Sx*y - 4xy» - 2y*, 4zy* - 3x*y - 2y\ -2y* - 3x*y - 4xy*. 

16. a; - sin-*, y, x = ^-|. 

16. a; - ± jV& 2 - y\ y - ± ^Va s - a* 

17. x = (a* - y*)», y - (a* - x*)». 

18. x = (a 1 - y 1 )*, y = (a 1 - a; 1 ) 1 . 

„. , x — 2o / — 

19. y = ± — - — V ox. 

20. a? - ± lV-2y*±Wy* + 256, y - ± ?\/4 - a:*. 



21. a? - 1 ± iV3 + 2y - y*, y - 1 ± 2\/2x - x*. 

22. a; - ±*-i^\/l6 - y\ 23. i - *-*' cos 2*. 

24. *> = ± 7= ,0 - i cos"" 1 — 2 - 
Vcos 20 «> 2 

25 . ^q cos ^ _, ( r 2a ± V4a« + <p* 



= cos" 1 y- 



sin* \ <p / 

26. <p - ± aVtan 0(3 - 4 sin* 0). 27. x = ^> y - a»x. 
28. x - jV2(y + Vl - y»), 2/ - iV2(* - Vl - **). 

Page 226. Art. 171. 

2. 4, -i. 3. 2, 2x - y + 1 - 0, x + 2y - 7 - 0. 
4.6*. 18*'. 6.2. l.~±. S.f. t-nj^T? 



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332 ANALYTIC GEOMETRY 

ID. 3xj* + 4*i. 1L ftri* - 8*1 + 6. 12.-^. IS. =*. 14.-. 16. 

2yi 9yi y x 

- fo j 1)t 16 - 2» - 2y + p - 0, 2s + 2y + p - 0. 17. -0.4364, 
0.4364. 18. 73° 44.4'. 19. 120°. 

Paget 131-233. Art. 181. 



1.^-te. 

dx 


8. ^ - 20s«. 
dx 


a**- 28 ** 


. <fy 2a 
ax 3** 


5 <fr ,. 9 

* <te 16*** 


*. * « 2 , 

* dx 3 


7 ^?- > 
dx x* 


ax 5x** 


ft * - 2 


10.*. • 

dx 3** 


dx 2 




"••s-* 


dt fit* 


16. * - }t*. 


16. J - 4x« + «x. 


if.J-a.-a. 


18. ^ - 3x« - Jx* + 3. 


19 d„.3x. + l. 
dx 2x* 


»*-*+Jr 


21. J - 6(2x + 1)«. 


22. * - 24*<3*» + 2)« - 2. 


as ^ _ * -a 


^ dy m 4x - 7^ 


dx V2x + 3 


dx 2 V2x» - 7* 


M dy = 2x + 7 


»-t --!-.• 


dx 3^0e* + 7x - 2)» 


27 ^ - - 1. 

" dx x 4 


dx x* 


29 * - 2 • 


,0 * . 1 . 


* ¥ * dx (x + 1)* 


dx 2\/(x + 1)» 


tt4 dy 20x 

31 " dx " (x - 1)»* 


22 *«= 2 


M " dx (x + 1)« 




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ANSWERS 333 



»*..'■ k.*- 1 — 



*• dx (x - 3)»* •** dx (*• + 1)«* 

36. P - 21x« - 24s* + 12a* 36. ^ * 



dx <fc 2\/x + 1 2\/x - 1 

87 *U » 0s» + 14x - 3 

# dx * 2 V3x» + 7s» - 3x + 2 
33 dy « 2ox + b _ 1 

" dx " 2\/ax l + bx + 6 2\/* + d 
^ dy 4x 4 + 10x ^ dy 2x 

TO - dx" -^JT+5 • ^ * " <* - 1)» 

41 *. \ 4f. *t . - «' 

' * <& ( x + l)Vx« - 1* dx V(x» - a»)« 

43. *■■ ^= + -37=1=. 44. *- -L-i + l*.. 

45. jjj-»5x< -3x* + 2x-2. 

46. ^- (x + a)« fc >(* - 6) w ->(w« + nx + dm - 6n). 

4t. ^- (x + l)*(2x - 1)*(16* + 1). 

43. dy _ 2-4* 4ft d« n*»-> 



50 



dx (x - 1)** w# dt (1 + 0» +1# 

dt 2« 



dt «* + 1) VF^l 
51. 0,3, 12. 63. 1:4, 1:8, 1:16. 

54. x - y - 6 *= 0, * + y + 6 « 0; 

29x - y - 38 - 0, x + 29y - 582 - 0. 

55. At the points whose abscissas are r * 

o 

M Xi* — 1, x Xi 1 , . 

66. y - yi - ^ (x - x,), y - y, - j _ — (a; - Xi). 

57. (-1, -6), 7x + y + 13 =0. 

58. At (1, 0) at 135 % , at (-3, -4) at 18* 26'. 
60. 1.0026026. 

Page 234. Art. 182. 

1 *£ - 5? q dy 3x» 4- 1 

x " dx * y«* *' dx 3y* + l" 

# dy 6%c . dp p dv v 

dx a*y dv v dp p 



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334 ANALYTIC GEOMETRY 

k *M _ 4x» - 8xy», dx _ 
da? 8x*y — 3y* ' dy ™ 
dy jr| dx x^ 

dx* x i' dy~ ""y** 

dy x +Va? 1 - y» 



e, 



i-- 



8x«y 


-3y» 






4x* - 


-8xy* 

7 * 

T# dx 


*♦ 
"F 


> 


*«(t> - 


■W 





y 

dp gy — 2ab — pp* d» _ ______ . 

dti m v»(v — 6) ' dp = o» — 2a& — p** # 

10. 3x + 4y - 26 - 0, Ax - 3y = 0. 

11. x - 6y + 17 - 0, Ox + y - 9 - 0. 

12. 8x + 5\/6y - 36 = 0, 25x - sV&y - 18 - a 
19. xVyi + y Vxi - Vasiyi - 0, 

xVii - y Vyi - XiVxi + 2/iVyi - 0. 

Page 237. Art. 185. 

L Rising for all values of x. 2. Rising for x > — 2, falling for x < — 2, 

3. Rising for x>0, never falling. 4. Rising for x>0, falling for x<0. 

5. Rising for all values of x, except x = 0. 6. Falling for all values of x 

except x = 0. 7. Rising for x > V3, and x < — V3, falling for 

- V3 < x < V3. 8. Rising for x > 1+ 8 and * < "1 V , 

falling for 1 ~ 3 < a? < 1 + 8 • •• Rising for x >1 and x <J, falling 

for } < x < 1. 10. Rising for — 1 < x < 1, falling for x > 1 and x < — 1. 
11. Rising for — 1 < x < 1, falling f or x > 1 and x < — 1. 12. Rising for 
x>2 and x< — 1, falling for — Kx<2. 13. Rising for x>— 2, falling 
for x < — 2. 14. Rising for x > 1 and — Kx <0, falling for x < — 1 and 
<x <1. 15. 278, 19, 3, y decreasing twice as rapidly as x is increasing. 

2 ± V'22 
16. 5 no real values of x. 

Page 239. Art. 186. 

1. Min. at x = 0. 2. Min. at x = 2. 3. Max. at x - 3. 4. Max- 

at (0, 2), Min. at (0, -2). 5. Max. at x = }(1 - Vl3), Min. at 

x - |(1 +Vl3). 6. Max. at x - 0, Min. at x - 4}. 7. Min. at 

x - 16. 8. Min. at x - 3. 9. (4.5, 4.1). 10. v% * m * " . 

Pages 241, 242. Art. 187. 

1. Upward x>0, downward x<0, Infl. at x « 0. 2. Upward for all 
values. 3. Upward x>0, downward x<0, Infl. at x =* 0. 4. Upward 



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ANSWERS 335 

x <0, downward x >0, Infl. at x = 0. 5. Upward x > 1 and x < — 1, down- 
ward — Kx<l, Infl. at x = ±1. 6. Upward x>l, downward x<l, 
Infl. at x » 1. 7. Upward a?> J and x<0, downward 0<x<}, Infl. 
at x =» and x = |. 8. Upward x>$, downward x<4, Infl. at x = J. 
9. Upward x>J\/3 and x<-J\/3, downward -i\/3 <x < iVz, 

Infl. at x - ±i V3. 10. Upward x> 4 + 3 V ^ and x < 4 " j^*' 

downward 4 "/ 1 "^ x < 4 + /", Infl. at x = 4 * ^. 11. 
3 3 3 

-2, Max. at x = 1 - J\/6, Min. at x = 1 + }\/6. 12. 0, 2, 20 in. 

per sec. 13. No points, 1, 2. 



(6s + 2)dx. 

9xdx 

W 





Pages 246, 246. Art. 190. 


5. -0.15,0.25. 


6. -0.09428, 0.09428. 7. 


8. (3x» + 4)dx. 


9. (4x» - 9s* + 4x)ox. 10. 


11 *** . 


12. -^dx. 18 


Vx* + 4 


xt 


u *** . 


15. 0*o7, 64. 


V(x* + 5)« 




Pages 249, 260. Art. 198. 


11. x* = iy. 


12. x' - Zy -f 



18. 2x ! - 3y + 4(3 - V2) « 0. 14. Af V2 square units. 

15. V\/6 square units. 17. 64 square units. 

18. 20} square units. 19. 25$ square units. 

20. 3V2 square units. 21. J square units. 
22. 2} square units. 

Pages 253, 254. Art. 197. 

1. / ^ 3 cos 3x. 2. / - sin 2x. 

ax ax 

8. ^ - -2 sin (2x + 1). 4. ^ = cos 2x. 

5. ~ = 3 cos 3x cos 2x — 2 sin 3x sin 2x. 
ax 

6. P = 3 sec' 3x. 7, ^ = sec* x. 
ax ax 

8. ,- = 15 tan* 5x sec 2 5x. . 9. ~ = x cos x + sin x. 

ax ax 



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336 ANALYTIC GEOMETRY 

10. ^ - (3x* + 2x) cos (x« + X*). • 

11. ^ - (2x + 3) cos (x« + 3a: - 4). 

15. g| - -3 sin (fix + 4). IS. ^ - } cot 3xV«n3x. 

14. ^ - } sin 4a:. 15. g - J tan xVsec x(3 cos* x + 1). 

16. -g — — mng(cot* +1 gx + cot*"" 1 qx). 

17 dy = 2 sin a?. 
* da: (1 + cos x)* 

d0 1 — sin 30 d$ 

20. & - 3 cos x(l - 4 sin* x). 21. ^ - cot i*. 

ax ax 

22. 2 square units. 23. 0.7071, -0.4161. 

24. 1,0, oo. 

25. Max. at x « (4n + 1) =» Min. at x «■ (4x + 3) ~» Int. at x » nr. 

26. -i cos 3x + C. 2(7. -f cos (3x - 1) + C. 

28. i sin 4x + C. 29. i sin (4x - 2) + C. 30. J sin* x + C. 
31. i sin 4 x + C. 32. -i cos* x + C. 

33. — r^r sin* +1 x + C. 34. y = sin x. 

n + 1 



Paget 259, 260. Art. 203. 

dy 2x-H7 . (2x + 7)dx 

*• dx " x* + 7x' y * x* + 7x 



2 dy _ 0.8686 ? rf _ 0.8686dx 
dx ™ x ' x 

a dy = _1 , ^ _dx 
dx ~~ x x 

4 dy = 0.8686 t rf _ 0.8686dx 
dx x x 

6. ^ - 2xe* f , dy - 2xe**dx. 

7. ^ - 6xe** ,+ <, dy - 6xe** ,+ «dx. 



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ANSWERS 



337 



8. 

9. 
10. 
11. 

12. 
13. 
14. 
16. 
16. 

17. 
18. 
19. 



-X — e*(sin x + cos x), dy 



j2 - 2a* log. a, dy 



e*(sin x + cos x)dx. 
2a te log« adx. 



4.6052xl0* + «, dy - 4.6052xl0*+«dx. 



dy 
dx * 

^ - 3x(3x - 2)'-> + (3x - 2)- log (3x - 2), 

<fy = [3x(3x - 2)'-i + (3x - 2)* log (3* - 2)]dx. 



*-*<*-«->,* 



|(e» - e-*)dx. 



d< 



— afce-«<, di — — a&e-«*d£ 



di RI -£! .. £/ _* 

3i" ~T« L '*" "T 6 Ldt 



e""*(cos « — sin x)dx. 



3^ = e~*(cos x — sin a;), < 
ax 

j - -J«-*«(6 sin 2* + cos 2/), 

dt _ -i<H«(6 sin 2* + cos 2t)dt. 
dy (x + 1)» (x + l)%fe 

dx = x* + 1 y x« + 1 
dy 
dx 

^ - (3 - 4* - tte*)«-» , l dy - (3 - 4x - 6x')e-*'dx. 



(4 + ?) (2x + log x), dy - (4 + D (2x + log x)dx. 



dy 



2(x» + l) to+ * 



dx 

dy - 2(x* + l)** + » 



2xM-_3x 
. x* + 1 
2x» + 3x 



21. 
24. 
26. 
27. 
29. 
81. 

88. 

85. 

87. 
39. 



x* + 1 

1, 7.39. 22. 0.4343, 0.0434 

No Max. point, Min. (i, 1.193). 
2.3026 square units. 
6.693 square units. 
3 log Cx. 
log C sin x, 



+ log (x* + 1) 
+ log (x* + 1) 



23. 



dx. 
(1, 0.6931). 



28. log C(x - 1). 
30. x + log Cx. 
32. ie** + C. 



ix* - 2x + log Cx. 

a 8 * 
3ToJ^ + a 

y* - e» f . 

22 



84. 



»-ta«^ + i-Si- 



36. x - 4<r* + C. 
38. y = e»*. 



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338 ANALYTIC GEOMETRY 

Paget 266, 266. Art 208. 

L 11. 2. 13. 8. V61, V85, Vl66. 

9. (3, 4, 0). 10. (37, -36, 23). 1L 2: -1. 

la. (i, -l, -9). 

Page 269. Art 210 

1. * VV* \/3, i \/3, or - iV3, - * \/5. - i V%' 1, 1, 1. 

2. A Vio, - a Via - i VTo, or - ^ Vio, A Vio 

iVI6;7, -4, -5. 

3.?,*, -*,or -$, ~M;2,3, -6. 

4. -M, ~f,orf, -|,f. 

5. A, - A, A, or - A, A, ~ A- 

e. A, A, -tt,or-A, -A,B. 

7. «,A, -H,or-«, -*,«. 

8. - 2, -9, 6. 9. (1, 0, 0), (0, 1, 0), (0, 0, 1). 
10. (1, 0, 0,) (0, cos 0, cos y). 12. 45° or 135°. 

18. 48° 51' or 131° 9'. 14. J\/2, i\/2, 0. 

15. J V3, J, 0. 

Page 273. Art 213. 

1. (2, 60°, 45?, 120°) or (-2, 120°, 135°, 60°). 

2. (8, 60°, 120°, 45°) or (-8, 120°, 60°, 135°). 

3. ( V3, 54° 44', 54° 44', 54° 44') or 
(- VS, 125° 16', 125° 16', 125° 16'). 

4. 45° or 135°. 5. (8, 60°, 30°) or (-8, 240\ 150°). 

6. (4, 210°, 120°) or (-4, 30°, 60°). 

7. (4, 135°, 60°) or (-4, 315°, 120°). 

8. 77° 56.6'. 9. f , - i, J, or -»,!,- f. 

10. *. 11. -*?. 12. -f. 

Paget 277, 278. Art 218. 

U = 3. z « -4. 2. z «■ 0. y = 0. * « 0. 

3. y s + z* - 9. x* + y* - 16. 4. 8s + 4y - 10* - 15 - 0. 

5. y - z = 0. 6. 3* - y* = 0. 

7. x* - y 8 - z* - 0. 8. s* - 4x + 8y - 6* + 29 - 0. 

9. x* + y* + z* - 3* - 4* - 0. 

10. x» + y* + s 8 - 6x - 72 = 0, x» + y* + ** - 2x - 80 = 0. 

11. 17x« + 17y* + 17 2 « - 90s - 25Sy + 265 - 0. 
** + y % + z* - 2x - 2y - 47 - 0. 

12. x* + y* + « l - 2x + 2y - 4* - 75 = 0. 

13. x* - y* - z» = 0. 14. s» - y» + «» - a 

22 



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ANSWERS 339 

16. x* - y* - 2* = 0. 16. x - y* - z* - 0. 

17. x* - 4y* - 42* - 0. 18. x* + 2/* - 2* = 0. 

19. x 4 + y 4 + s 4 + 2x V + 2xV + 2y V - 4y* - 4** * 0. 

20. s* + y* + 2* - 2z - 0. 21. x» + y* - sin 1 * = 0. 
22. x* - (sin-ty)* + 2* - 0. 

»• *! + £ + £ - l- * 4 - -! + r! + -i * L 

a 1 6* 6* a* b* a* 

25. (x* + y* + «* + 12)* - 64(y* + z*) - 0. 

Pages 281, 282. Art 221. 

4. 3y* - 5** + 2 = 0, 3x« - 22» - 10 - 0, 5x* - 2y* - 18 - 0. 

5. 3y - z % - 0, 3x* - 2* - 0, x* - y - 0. 

6. z 4 + a*y* - oV = 0, 2* + ax - a* = 0, x t + y* - ox « 0. 

7. y s - z* - 0, x' + 2* - a* - 0, x' + y' - a* =0. 

8. m*y* + 2* - a*m* - 0, 2 - mx = 0, x s + y t - a* = 0. 

9. 2 s - 3y* = 0, 2* - 3ox - 0, y s - ox - 0. 

Pages 291. Art. 229. 

21. x s + y* 4- 22* - 4. 24. x* + 2» - 2px + p* - 0. 



1. 



8. 





Page 297, 298. Art. 287. 


1 2 
3*-3 y - 


- 3 2 3 -u, 4 + _ 2 -h_ 2 - 


2 _. 1 

S' + I^" 


, 2 „_* -n *-l^4. JL - 


4 _L 7 

9 X + 9 y - 


4 ,1 A ^ I ^ 1* 

•9 2 + 3 =0 '^1 + ^ + i 


12 1 


,12 18 „ x , y . 


T7*-T7 y 


+ r7*-r7-°-i + -i8 + 



1. 



5. 2x - 3y + 22 - 1 - 0. 6. 3x + 2y - 2 - 4 = 0. 

7. x - 2y ± 22 - 15 = 0. 8. 6x - 3y + 2 - 2 = 0. 

9. x + 2y - 22 - 6 - 0, 91x - 122y + 462 - 318 - 0. 
10. 3x - y - 4z - 1 = 0. 
12. 3x - 4y + 22 - 4 - 0. 
14. 3x - 4y + 42 - 16 - 0. 
16. 3x - 4y + 22 + 29 = 0. 
18. 56° 15' or 123° 45'. 

20. 67° 7' or 112° 53'. 

21. 5x - lly - 82 + 14 - 0, 23x + 25y - 2O2 - 28 - 0. 

22. x - 2y - 22 + 5 = 0, 4x + y + 2 - 7 - 0. 

23. x + 2y - 52 - 5 - 0, llx - 8y - 2 - 13 = 0. 

24. ±6. 25. (2, -1, 2). 26. 31° 1', 64° 37', 73° 24'. 
27. 16° 36', 25° 23', 58° 59'. 



11. 


x + y + 2-3 = 0. 


13. 


x-2y + 2-1=0. 


15. 


3y - 42 + 5 - 0. 


17. 


I. 


19. 


48° 11' or 131° 49'. 



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340 ANALYTIC GEOMETRY 

Paces MI-MS. Art Ml. 

1. (1, 1, 0), (2, 0, -3), (0, 2, 3). 

5. (2. -3, 0), (8, 0, -3), (0, -4, 1). 
t. (3, -2,0), (1,0, -1), (0, 1, -|). 
4. (-1,1,0), (-1,0,3), (0,2, -2). 
«. (1, J, 0), (2, 0, 1), (0, 1, -1). 

6. (1) * + y - 2 - 0, 3* + z - 3 - 0. 
(2) * - 2» - 8 - 0, * + 2* - 2 - 0. 

7. (3) * + y - 1 - 0, y + 2z + 2 - 0. 
(4) 2* - - ' - - - - 



y + 2-0, 5y + 2« - 8 « 0. 

g — ■ 
' 3 
«-l 



"• W -1 1 3 



9. 11* + 5y - 3 - 0, 3* - 5* + 11 - 0. 

10. 2x + 5y - 1 - 0, 2x - 5s + 14 - 0. 

11. * - 2 - 0, y + 2s + 5 - 0. 

12. x — 2 « 0, y - 5 - 0. 

15. x + 3y + 8 - 0, 2x - 3s + 10 = 0. 

14. x - 3 - 0, y + 1 - 0. 15. * - 2 - 0, s + 3y - 0. 

16. x + y - 2 - 0, 3s + z - 11 - 0. 

17. 2x + y - 5 - 0, x - s - 1 - 0. 

18. Zx - 2y - 0, 2* - s « 0. 19. \. 20. 0. 22. i, -}, }. 
28. x - y + 2s - 7 - 0. 24. 7a? - 12y + s - 14 - 0. 

25. x + y + s - 3 « 0. 26. 2s - y + * - 1 - 0. 

27. 2s -3y- 4* + 6-0. 28. '* + * = ^- 2 = i + i 



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\ 



INDEX 

Numbers refer to pages. 



Abscissa, 13, 14 
Algebra, formulas of, 2 
Algebraic, equations, 154 

hyperbolic type, 159, 197 

parabolic type, 158, 195 
Algebraic functions, 225 
Amplitude, of function, 169 

factor, 169 
Analytic, geometry, 1 

methods, 39 
Anchor ring, 277 
Angle of lag, 172 
Angles, bisectors of, 70, 297 

definition, 24 

direction, 267 

formed by lines, 24, 271 

formed by planes, 295 

tangent of, 27 

vectorial, 30 
Applications, of ellipse, 131 

exponential functions, 164 

hyperbola, 150 

parabola, 113 

straight line, 79 
Arch, parabolic, 113 

elliptic, 131 
Areas, by integration, 247, 252, 
254, 259 

of polygon, 38 

of triangle, 36 
Asymptotes, 139 
Axes, coordinate, 12, 261 

polar, 30 

rotation of, 34 

translation of, 33 



Axis, conjugate of hyperbola, 137 
major, of ellipse, 120 
minor, of ellipse, 120 
of parabola, 102 
of symmetry, 50 
transverse, of hyperbola, 137 

B 

Bisector, of angles formed by lines, 

70 
of angles formed by planes, 

297 
Boyle's law, 150 



Cardioid, 185, 187 
Cassinian oval, 163 
Catenary, 167 
Circle, equation, 86, 87, 96, 279 

imaginary, 87 

locus problems, 94 

point or null, 87 

radical axis, 94 

satisfying three conditions, 88 

systems, 92 
Cissoid of Diodes, 160 
Concavity, 236, 239 
Conchoid of Nicomedes, 162 
Condition second degree equation 
represents two straight 
lines, 77 
Conditions for locus, 45 
Cone, 290 
Confocal ellipses, 132 



341 



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342 



INDEX 



Conic sections, 98 
Conicoids, 284 
Conies, definition of, 99 

degenerate, 157 

diameters of, 210, 214 

directrix of, 99 

eccentricity of, 99 

focus of, 99 
Conjugate, axis of hyperbola, 137 

diameters, 212, 213 

hyperbolas, 141 
Continuous, functions, 235 
Coordinates, axes of, 12 

cartesian, 12 

oblique cartesian, 15 

origin of, 9 

polar, 12, 29, 269 

rectangular in plane, 13 

rectangular in space, 261 

relation between, 32 

spherical, 270 

transformation of, 33 
Constants, 44 

arbitrary, 53 
Construction of, ellipse, 129 

hyperbola, 148 

parabola, 112 
Curves, concavity of, 236, 239 

cycle of, 169 

empirical, 188 

falling, 236 

in space, 278 

maximum point of, 235, 237 

minimum point of, 235, 237 

normal to, 223 

periodic, 168 

points of inflection, 236, 239 

probability, 199 

projection of, 280 

proper sine, 168 

properties of, 235 

rising, 236 



Curves, sine, 167 
slope of, 222 
tangent to, 222 

Cycle of curve, 169 

Cycloid, 182 

Cylindrical surfaces, 274 



Degenerate forms of conies, 157 
Derivatives, 222 

of algebraic functions, 226 

of exponential functions, 254 

of logarithmic functions, 254 

of trigonometric functions, 
250 
Descartes, 1 
Diameters, conjugate, 212, 213 

length, 212, 213 

of conic, 210 

of ellipse, 210 

of hyperbola, 213 

of parabola, 213 
Differential triangle, 245 
Differentials, 242 

definition of, 243 
Differentiation, 225 

fundamental method, 225 

of implicit functions, 233 
Direction angles, 267 

cosines, 267 
Directrices of, conic, 99 

ellipse, 117, 120 

hyperbola, 134, 137 

parabola, 100 
Discriminant, 2 
Discussion of equations, 48, 174, 

282 
Distance, between two points, 17, 
19,263 

from point to line, 68 

from point to plane, 296 



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INDEX 



343 



Division, external, 19 
internal, 19 
of line segment, 19, 23, 264 



Eccentricity of, conic, 99 

ellipse, 100, 118 

hyperbola, 100, 136 

parabola, 100 
Ellipse, applications, 131 

center, 120 

conjugate diameters, 212 

definition, 100, 117 

diameters, 210 

directrix, 117, 120 

eccentricity, 100, 118 

equation, 118, 121, 124, 129 

equation of tangent to, 234 

focus, 117, 120 

general equation, 128, 155 

imaginary, 123 

latus rectum, 120 

major axis, 120 

minor axis, 120 

point, 123 

sum of focal distances con- 
stant, 130 

vertices, 120 
Ellipsoid, 284 
Elliptic paraboloid, 288 
Empirical curves, 188 
Epicycloid, 185 
Equations, algebraic, 154 

discussion of, 48, 174, 282 

exponential, 163 

exponential type, 197 

general, of second degree, 87, 
106, 124, 144, 154 

graph of, 46 

hyperbolic type, 159, 197 

linear, 65, 19) 



Equations, locus of, 45 
logarithmic, 165 
of circle, 86, 87. 96 
of curves in space. 278 
of ellipse, 118, 121, 124, 129 
of hyperbola, 135, 138, 142, 

143, 144, 147 
of line, 59, 62, 63, 64, 65, 78 
of parabola, 100, 102, 104, 111 
parabolic type, 158, 195 
parametric, 180 
plotting, 46 
polar, 175 

transcendental, 154, 163 
trigonometric, 167 

Exponential, equations, 163 
type, 197 

Extent, 51 



Focus of, conic, 99 

ellipse, 117, 120 

hyperbola, 134, 137 

parabola, 100 
Formulas, algebraic, 2 

differentiation, 230, 252, 255 

integration, 248, 253, 258 

logarithmic, 2 

summary of, 303 

trigonometric, 3 
Frequency, 172 
Functions, 216 

algebraic, 225 

amplitude of, 169 

continuous, 235 

decreasing, 236 

explicit, 218 

implicit, 218 

increasing, 236 

maximum value, 235 

minimum value, 236 



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344 



INDEX 



Functions, of variables, 44 
period of, 169 
periodic, 168 
quadratic, 107 
single-valued, 235 
trigonometric, 7 



Graph of equation, 46 



Harmonic, conjugates, 206 

motion, 170 

ratio, 206 
Higher plane curves, 154 
Hooke's law, 80, 193 
Hyperbola, applications, 150 

asymptotes, 139 

center, 137 

conjugate, 141 

conjugate axis, 137 

conjugate diameters, 213 

definition, 100, 134 

difference between focal dis- 
tances constant, 148 

directrix, 134, 137 

eccentricity, 100, 135 

equation, 135, 138, 143, 147 

equation of tangent to, 234 

equilateral, 142, 150 

focus, 134, 137 

general equation, 146, 155 

latus rectum, 137 

principal axis, 137 

rectangular, 142 

transverse axis, 137 

vertices, 137 
Hyperbolic, paraboloid, 289 

type, 159, 197 
Hyperboloids, 285 
Hypocycloid, 183 



Imaginary, circle, 87 

ellipse, 123 

number, 47 
Inclination of line, 25 
Increments, 216 
Infinite, variable becomes, 221 
Initial line, 30 
Integral, indefinite, 247 
Integration, 246 

constant of, 247 

formulas, 248, 253, 258 

methods of, 248 
Intercepts, 49, 283 
Involute of circle, 186 



Latus rectum of, ellipse, 120 

hyperbola, 137 

parabola, 102 
Least squares, 193 
Lemniscate, 163 
Iimacpns of Pascal, 179 
Limits, 220 

theorems of, 221 
Linear equations, 65, 191 
Lines, applications, 79 

directed, 8 

direction cosines, 267 

general equation, 65 

inclination, 25 

initial, 30 

in polar coordinates, 78 

in space, 298 

intercept equation, 63 

normal equation, 64 

parallel, 27 

perpendicular, 27 

point direction equation, 299 

point slope equation, 59 



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INDEX 



345 



Lines, polar equation, 78 

projection equation, 298 

slope, 25, 65 

slope intercept equation, 61 

systems, 71 

two point equation, 62, 300 
line segment, 9 

addition and subtraction of, 
10, 11 

division of, 19, 23, 264 

magnitude of, 9 

numerical value of, 9 

projection of, 266 

value of, 9, 16 
Iituus, 180 
Loci, algebraic, 154 

composite, 53, 77 

in space, 274 

of equations, 45, 55 

of points, 45 

of polar equations, 175 

through intersection of loci, 
75 

transcendental, 154 
Logarithmic, equations, 165 

paper, 200 
Logarithms, formulas of, 2 

M 

Major axis of ellipse, 120 
Maximum, 235, 237 

test of, 238 
Method of least squares, 193 
Minimum, 235, 237 

test of, 238 
Minor axis of ellipse, 120 

N 

Newton's law of cooling, 167 
Normals, 223, 292 



Null circle, 87 
Number, imaginary, 47 

O 

Oblate spheroid, 285 
Oblique coordinates, 15 
Ordinate, 13, 14 
Origin of codrdinates, 9, 29 
Orthogonal projection, 266 



Parabola, applications, 113 

axis, 102 

construction, 112 

cubical, 158 

definition, 100 

diameters, 213 

directrix, 100 

eccentricity, 100 

equation, 100, 102, 104, 111 

equation of tangent to, 234 

focus, 100 

general equation, 109, 155 

latus rectum, 102 

semi cubical, 158 

vertex, 102 
Parabolic type, 158, 195 
Paraboloids, 288 
Parameter, 180 
Period of function, 169 
Periodic, curve, 168 

function, 168 
Periodicity factor, 169 
Phase angle, 174 
Planes, angle between, 295 

determinant equation , 292 

equations, 292 

general equation, 292 

in space, 292 

intercept equation, 294 



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346 



INDEX 



Planes, normal equation, 293 

normal to, 292 

parallel to axes, 274 
Point, circle, 87 

ellipse, 123 

imaginary, 47 

initial, 8 

locus of, 45 

of inflection, 239 

polar coordinates of, 12, 29, 
269 

rectangular coordinates of, 12, 
261 

spherical coordinates of, 270 

terminal, 8 
Polar, coordinates, 12, 29, 174* 269 

equation of circle, 96 

equation of ellipse, 129 

equation of hyperbola, 147 

equation of line, 78 

equation of parabola, 111 
Pole, definition, 29 
Poles and polars, 206 

properties of, 209 
Powers of e, 5 
Probability curve, 199 
Projectile, path of, 114 
Projection, orthogonal, 266 

of curves, 280 
Prolate spheroid, 285 
Proper sine curve, 168 

Q 

Quadratic function, 107 
Quadrie surfaces, 284 

R 

Radical axis, 94 
Radius vector, 30 
Rectangular system, 13 



Reflector, 115 

Revolution, surfaces of, 276 

Rotation of axes, 34, 109, 127, 156 

formula "for, 155 
Ruled surfaces, 286, 290, 291 

S 

Simple harmonic motion, 170 
Sine curve, 168 
Slope, definition, 25 

formula for, 25 
Spheres, 276 

Spherical coordinates, 270 
Spheroids, 285 
Spirals, 178 

Archimedes, 179 

center of, 178 

hyperbolic, 179 

logarithmic, 178 

parabolic, 179 
Supplemental chords, 215 
Surfaces, 274 

cylindrical, 274 

equations of, 274 

of revolution, 276 

quadrie, 284 

ruled, 286, 290, 291 

sections of, 279 

trace, 279 
Symmetry, 49, 175, 177, 282 

algebraic properties, 50 

axis of, 50 

center of. 49 



Tables, e* and e"*, 5 

of logarithms, 308 

of trigonometric functions,310 
Tangents, 223 

equations of, to conies, 234 
Torus, 277 



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INDEX -*\_ 347 

Trace of surfaces, 279, 283 V 

Transcendental equations, 154, 163 

Transformation of coordinates, 33 Value of ]ine ^^^ 9> 16 

Translation of axes, 33, 107, 123, Variables, 44, 216 

143,157 become infinite, 221 

Transverse axis of hyperbola, 137 dependant, 46 

Triangle, area of, 36, 39 functions of, 44 

center of gravity, 40 independant, 46 

Trigonometric equations, 167 Vectorial angles, 30 

Trigonometry, formulas of, 3 Vertice8 o{> ^ m 

functions of, 7 hyperbola, 137 

P 0001 ^ 183 parabola, 102 

Trumpet, 180 

U w 

Uniform circular motion, 170 Witch of Agnesi, 162 



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